Inverse relations found by switching x and y, verifying a pair of inverses with both compositions, the horizontal line test and restricting a domain, inverses of cubic and higher power functions, and inverting a model to solve for the other variable.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 6 — Rational Exponents and Radical Functions
Use Inverse Functions
Objectives
Five outcomes. The second one is the definition; everything else is how to meet it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 438-445 — the lesson these objectives are drawn from
Warm-up
Lesson 6.3 ended with a puzzle: f of f of x came out as x. That is not a coincidence worth ignoring.
Discussion prompt
Take f of x equals 3x minus 5 and describe in words what it does to an input. Then describe, in words, how you would get back to the original input from the output.
Hint: Reverse the operations, in reverse order.
Answer:
Forwards: multiply by 3, then subtract 5. Backwards: add 5, then divide by 3.
\[ f^{-1}(x) = \tfrac{x+5}{3} = \tfrac{1}{3}x + \tfrac{5}{3} \]
Reversing the operations and reversing their order is the whole idea, and Lesson 6.3's composition is what makes it precise: two functions are inverses when composing them either way returns the input untouched.
Concept
An inverse relation swaps the inputs and outputs of the original, so its graph is the reflection of the original's in the line y equals x. When both the relation and its inverse are functions, they are called inverse functions, and each undoes the other.
inverse function — A function g is the inverse of f when f of g of x equals x and g of f of x equals x for every x in the appropriate domains. It is written f to the negative 1, which is not an exponent.
\[ f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x \]
The notation f to the negative 1 is unfortunate but standard. It does not mean one over f: for f of x equal to 3x minus 5, the inverse is not 1 over 3x minus 5.
Figure (svg): Two columns contrasting a function's operations with the reversed operations of its inverse
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 438-438
Section
Section 1
Concept
An inverse relation interchanges the input and output values, so its domain and range are the original's range and domain. To find its equation, swap x and y in the original and solve the result for y.
inverse relation — The relation obtained by interchanging the input and output values of a relation. Its graph is the reflection of the original's in the line y equals x.
\[ y = 3x-5 \;\Longrightarrow\; x = 3y-5 \;\Longrightarrow\; y = \tfrac{1}{3}x + \tfrac{5}{3} \]
The reflection in the line y equals x is exactly what swapping the coordinates does to every point: the point with coordinates a and b becomes the point with coordinates b and a.
Figure (svg): A relation and its inverse shown as two tables and as two graphs reflected in the line y equals x
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 438-438 — Find an inverse relation
Picture it
Example 1: a linear relation and its inverse.
Figure (svg): A relation and its inverse shown as two tables and as two graphs reflected in the line y equals x
The two tables have their rows exchanged and the two graphs are mirror images across the dashed line. Both descriptions say the same thing.
Worked example
Example 1. Three lines.
\[ \text{Find an equation for the inverse of } y = 3x - 5. \]
Write the original relation
Why: It is already solved for y.
\[ y = 3 x - 5 \]
Switch x and y
Why: Every x becomes a y and every y an x.
\[ x = 3 y - 5 \]
Isolate the y term
Why: Adding 5 to both sides.
\[ x + 5 = 3 y \]
Solve for y
Why: Dividing by 3 gives the inverse relation.
\[ y = (\frac{1}{3}) x + \frac{5}{3} \]
Figure (svg): The solution to Worked example find an inverse relation shown as a ladder of expressions, one row per algebraic move
\[ y = \tfrac{1}{3}x + \tfrac{5}{3} \]
Verify: check a point and its swap
Why: The original passes through (2, 1), since 6 minus 5 is 1. The inverse should therefore pass through (1, 2): one third plus five thirds is two. It does. Every point of one graph has its coordinates reversed on the other, which is what reflecting in y equals x means.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 438-438
Matching
Swap and solve.
Match the pairs
Why: Each inverse's slope is the reciprocal of the original's, which is what reflecting a line in y equals x always does. A slope of 3 becomes one third, and a slope of negative 3 becomes negative one third.
Worked example
Guided Practice 1 to 3.
\[ \text{Find the inverses of } f(x) = x+4, \; f(x) = 2x-1, \; f(x) = -3x+1. \]
First: swap and solve
Why: X equals y plus 4, so y equals x minus 4.
\[ x - 4 \]
Second: swap and solve
Why: X equals 2y minus 1, so 2y is x plus 1.
\[ \frac{x + 1}{2} \]
Third: swap and solve
Why: X equals negative 3y plus 1, so negative 3y is x minus 1.
\[ \frac{1 - x}{3} \]
Note the pattern
Why: Each inverse reverses the operations in reverse order.
Figure (svg): The solution to Worked example three more inverses shown as a ladder of expressions, one row per algebraic move
\[ x-4, \; \tfrac{x+1}{2}, \; \tfrac{1-x}{3} \]
Verify: check the third with a value
Why: F of 2 is negative 5, so the inverse should send negative 5 back to 2: 1 minus negative 5 is 6, over 3 is 2. It does. Note that the third inverse has a negative slope, matching the original — reflecting a line in y equals x replaces its slope by the reciprocal, and the reciprocal of negative 3 is negative one third.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 439-439
Trap
\[ f(x) = 3x - 5 \]
Interpret f to the negative 1 as a reciprocal
Why: The superscript is treated as an exponent, as it would be on a number.
\[ f^{-1}(x) = \frac{1}{3x-5} \quad \text{(wrong)} \]
Composing that with f gives 1 over the quantity 3 times 1 over 3x minus 5, minus 5 — nothing like x.
\[ f^{-1}(x) = \tfrac{1}{3}x + \tfrac{5}{3} \]
Read the notation as a name, not an operation
Why: F to the negative 1 names the function that undoes f, and nothing more.
\[ f^{-1}(x) \neq \frac{1}{f(x)} \]
The textbook flags this in a Reading note. The notation is entrenched and unfortunate; the only defence is to read it as a word rather than as arithmetic.
Fill the middle
Example 1, at the second step.
Fill in the blanks
y = 3x - 5 \;\Longrightarrow\; x = 3y - 5
Why: Switching gives x equals 3y minus 5, which is then solved for y. Doing the swap first and the solving second keeps the two steps separate; trying to do both at once is where the sign errors come from.
Sorting
The inverse's points have the coordinates reversed.
Sort into buckets
For y = 3x - 5 and its inverse, sort each point.
Any point where the two graphs meet must be on the line y equals x, because that is the only place a point equals its own reflection.
Prediction
Commit before reasoning.
Predict first
A function has domain all reals and range the non-negatives. What are its inverse relation's domain and range?
Correct: Domain the non-negatives, range all reals.
\[ \text{domain of } f^{-1} = \text{range of } f, \quad \text{range of } f^{-1} = \text{domain of } f \]
Why: Interchanging inputs and outputs interchanges the domain and the range, so what was the range becomes the domain. This is why the inverse of x squared has domain restricted to the non-negatives: the parabola's range was the non-negatives. Tracking the swap of these two sets is what makes domain restrictions in the next idea predictable rather than surprising.
Section
Section 2
Concept
Two functions f and g are inverses when f of g of x equals x and g of f of x equals x. Both orders must be checked, because a function can undo another in one direction without undoing it in the other.
\[ f(g(x)) = x \quad \text{and} \quad g(f(x)) = x \]
This is the one case from Lesson 6.3 where the two orders of composition agree, and agreeing on the identity function is precisely what makes the pair inverses.
Figure (svg): Both compositions of a function with its inverse simplifying to x
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 439-439 — Verify that functions are inverses
Picture it
Example 2: verifying a linear pair.
Figure (svg): Both compositions of a function with its inverse simplifying to x
Each side simplifies to x by cancelling exactly what the other function introduced. The multiplication by 3 cancels the division, and the plus 5 cancels the minus 5.
Worked example
Example 2, both directions.
\[ \text{Verify that } f(x) = 3x-5 \text{ and } f^{-1}(x) = \tfrac{1}{3}x + \tfrac{5}{3} \text{ are inverses.} \]
Compose f with the inverse
Why: Substitute the whole inverse into f: 3 times the quantity one third x plus five thirds, minus 5.
\[ x + 5 - 5 \]
Simplify
Why: The 3 cancels the one third and the plus 5 cancels the minus 5.
Compose the other way
Why: Substitute f into the inverse: one third of the quantity 3x minus 5, plus five thirds.
\[ x - \frac{5}{3} + \frac{5}{3} \]
Simplify
Why: Again everything cancels.
Figure (svg): The solution to Worked example verify a pair shown as a ladder of expressions, one row per algebraic move
\[ f(f^{-1}(x)) = f^{-1}(f(x)) = x \]
Verify: trace one input through both
Why: Start at 4: f of 4 is 7, and the inverse of 7 is seven thirds plus five thirds, which is 4. The input came back untouched. Doing this with a single number is a fast informal check; the algebra proves it for every input at once.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 439-439
Fill the middle
Example 2, step 1.
Fill in the blanks
3\left(\tfracx___x + \tfrac______\right) - 5 = x + 5 - 5 = ___
Why: Distributing the 3 gives x plus 5, and subtracting 5 leaves x. Each operation of f was undone by the matching operation of the inverse, which is why the cancellation is complete rather than partial.
Worked example
Guided Practice 8, with the verification done in full.
\[ \text{Verify that } f(x) = 2x^3 + 4 \text{ and } g(x) = \sqrt[3]{\tfrac{x-4}{2}} \text{ are inverses.} \]
Compose f with g
Why: Substituting g into f gives 2 times the cube of the cube root, plus 4.
\[ 2(\frac{x - 4}{2}) + 4 \]
Simplify
Why: The cube and the cube root cancel, then the 2 cancels and the 4 cancels.
Compose the other way
Why: Substituting f into g gives the cube root of the quantity 2x cubed plus 4, minus 4, over 2.
Simplify
Why: The 4s cancel and the 2s cancel, leaving the cube root of x cubed.
Figure (svg): The solution to Worked example verify a non-linear pair shown as a ladder of expressions, one row per algebraic move
\[ f(g(x)) = g(f(x)) = x \]
Verify: check that the cube root needed no restriction
Why: The cube root of x cubed is x for every real number, positive or negative, because the index 3 is odd — Lesson 6.2's rule. Had the power been even, absolute value would have appeared and the pair would not have been inverses without a domain restriction.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 441-441
Error analysis
A student verifies a claimed pair of inverses.
Annotate
On: \( f(x) = x^2, \; g(x) = \sqrt{x}: \; f(g(x)) = (\sqrt{x})^2 = x, \text{ so they are inverses} \)
Both directions are part of the definition. Restricting f to non-negative x makes the pair genuine inverses, which is exactly what the next idea does.
Sorting
Both compositions must give x.
Sort into buckets
Sort each pair.
The cube pair works on all reals while the square pair does not, and the reason is the parity of the index — the same fact that has organised this whole chapter.
Prediction
Commit before reasoning.
Predict first
If f of g of x equals x for every x, must g of f of x also equal x?
Correct: No — the squaring and square-root pair is a counterexample.
\[ (\sqrt{x})^2 = x \quad \text{but} \quad \sqrt{x^2} = \lvert x \rvert \]
Why: The square of the square root of x is x for every x where the square root exists, but the square root of x squared is the absolute value of x, which differs from x on the negatives. So one composition can succeed while the other fails, and the definition demands both. That is why the textbook's verification has two clearly labelled steps rather than one.
Ranking
Checking that two functions are inverses.
Put in order
Why: The two directions are separate checks and both are required. Step five is what catches the squaring case: the algebra may look fine until a negative input is tried, and a pair that works only on part of the domain needs that part stated.
Section
Section 3
Concept
Reflecting a graph in the line y equals x turns a horizontal line into a vertical one. So the inverse of f is a function exactly when no horizontal line meets the graph of f more than once — the horizontal line test.
\[ f(x) = x^2 \text{ fails}; \quad g(x) = x^3 \text{ passes} \]
When a function fails the test, restricting its domain can rescue the inverse. Cutting the parabola down to the non-negative half makes the square root its genuine inverse.
Figure (svg): The squaring and cubing functions with their reflections, showing that only one inverse is a function
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 440-440 — Horizontal line test
Picture it
The squaring function fails; the cubing function passes.
Figure (svg): The squaring and cubing functions with their reflections, showing that only one inverse is a function
The reflected parabola is a sideways curve failing the vertical line test, which is why the horizontal line test on the original is the right thing to check.
Worked example
Example 4. The restriction decides which square root to take.
\[ \text{Find the inverse of } f(x) = x^2 \text{ with } x \ge 0, \text{ and graph both.} \]
Replace f of x with y and switch
Why: The relation becomes x equals y squared.
\[ x = y ^{2} \]
Take square roots
Why: Both signs appear, so the relation is not yet a function.
\[ y = +- \sqrt{x} \]
Use the restriction to choose a sign
Why: The domain of f was the non-negatives, so the range of the inverse must be too.
State the inverse
Why: The inverse is the principal square root.
\[ f\text{ inverse } (x) = \sqrt{x} \]
Figure (svg): The solution to Worked example restrict a domain shown as a ladder of expressions, one row per algebraic move
\[ f^{-1}(x) = \sqrt{x} \]
Verify: check both compositions on the restricted domain
Why: For x at least zero, the square of the square root is x and the square root of the square is x as well, since x is not negative and the absolute value does nothing. Both directions succeed, which they did not before the restriction. Had the domain been x at most 0, the inverse would be the negative square root instead.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 440-440
Sorting
Apply the horizontal line test.
Sort into buckets
Sort each function.
Every even power fails and every odd power passes, which is the parity rule appearing once more — this time as a statement about invertibility.
Worked example
Guided Practice 5, plus the negative-half version.
\[ \text{Find the inverse of } f(x) = x^6 \text{ with } x \ge 0, \text{ and of } f(x) = x^2 \text{ with } x \le 0. \]
First: switch and take the sixth root
Why: X equals y to the sixth gives y equal to plus or minus the sixth root of x.
\[ +- x ^{\frac{1}{6}} \]
First: choose the sign
Why: The domain was the non-negatives, so the inverse's range is too.
\[ x ^{\frac{1}{6}} \]
Second: switch and take square roots
Why: X equals y squared gives plus or minus the square root.
\[ +- \sqrt{x} \]
Second: choose the other sign
Why: The domain was the non-positives, so the inverse's range must be too.
\[ -\sqrt{x} \]
Figure (svg): The solution to Worked example two more restricted powers shown as a ladder of expressions, one row per algebraic move
\[ f^{-1}(x) = x^{1/6}; \qquad f^{-1}(x) = -\sqrt{x} \]
Verify: check the second on a negative input
Why: F of negative 3 is 9, and the inverse of 9 should return negative 3: the negative square root of 9 is negative 3. It does. Choosing the wrong sign would return positive 3, which is not the input the function came from.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 441-441
Trap
\[ f(x) = x^2 \;\Longrightarrow\; x = y^2 \;\Longrightarrow\; y = \pm\sqrt{x} \]
Report the plus-or-minus as the inverse function
Why: Both roots are kept, since both satisfy the equation.
\[ f^{-1}(x) = \pm\sqrt{x} \quad \text{(a relation, not a function)} \]
At x equal to 9 this assigns both 3 and negative 3, so it fails the vertical line test and is not a function at all.
\[ f(x) = x^2, \; x \ge 0 \;\Longrightarrow\; f^{-1}(x) = \sqrt{x} \]
Restrict the original's domain, then choose the matching sign
Why: The inverse's range must equal the original's domain, which decides which root to keep.
\[ \text{domain } x \ge 0 \;\Longrightarrow\; \text{range of } f^{-1} \text{ is } y \ge 0 \]
Without a restriction the inverse is a perfectly good relation but not a function. Restricting is not a trick to make the answer work; it is what makes the question have a function as its answer.
Fill the middle
Example 4, at the restriction.
Fill in the blanks
f(x) = x^2, \; x \ge 0 \;\Longrightarrow\; f^+(x) = ___\sqrt___
Why: The domain of f was the non-negatives, so the range of the inverse must be the non-negatives too, which selects the positive square root. Restricting to x at most zero would select the negative one instead — the same original formula, two different inverses.
Prediction
Commit before reasoning.
Predict first
What does a horizontal line meeting f twice become after reflecting in y equals x?
Correct: A vertical line meeting the inverse twice.
\[ y = k \;\to\; x = k \quad \text{under reflection in } y = x \]
Why: Reflection in y equals x swaps the coordinates, so a horizontal line becomes a vertical one. If a horizontal line met f twice, the corresponding vertical line meets the reflection twice, and the reflection fails the vertical line test of Lesson 2.1. The horizontal line test is the vertical line test applied to the inverse before the inverse is drawn.
Comparison
Fill the blanks. The parity decides everything.
Comparison matrix
| Question | f(x) = x^2 | f(x) = x^3 |
|---|---|---|
| Horizontal line test | fails | passes |
| Inverse is a function? | not without a restriction | yes, everywhere |
| Inverse formula | sqrt(x), on x >= 0 | cube root of x |
| Domain of the inverse | x >= 0 | all real numbers |
An even power folds the negatives onto the positives, and folding is exactly what cannot be undone. An odd power never folds.
Section
Section 4
Concept
For a power function built from a coefficient, a power and a constant, switching x and y and solving means undoing the constant, then the coefficient, then the power — the reverse of the order in which they were applied.
\[ y = 2x^3+1 \;\Longrightarrow\; f^{-1}(x) = \sqrt[3]{\tfrac{x-1}{2}} \]
An odd power passes the horizontal line test automatically, so no restriction is needed. An even power always needs one.
Figure (svg): A cubic and its inverse graphed together, symmetric about the line y equals x
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 441-441 — Find the inverse of a cubic function
Picture it
Example 5: f of x equals 2x cubed plus 1.
Figure (svg): A cubic and its inverse graphed together, symmetric about the line y equals x
The two curves are reflections in the dashed line, and the inverse is defined for every real number because the cubic passes the horizontal line test.
Worked example
Example 5, with the test done first.
\[ \text{Determine whether the inverse of } f(x) = 2x^3+1 \text{ is a function, then find it.} \]
Apply the horizontal line test
Why: The graph rises throughout, so no horizontal line meets it twice.
Switch x and y
Why: The relation becomes x equals 2y cubed plus 1.
\[ x = 2 y ^{3} + 1 \]
Undo the constant, then the coefficient
Why: Subtracting 1 and dividing by 2 isolates y cubed.
\[ \frac{x - 1}{2} = y ^{3} \]
Undo the power
Why: Taking the cube root of both sides, with no sign choice needed.
Figure (svg): The solution to Worked example invert a cubic shown as a ladder of expressions, one row per algebraic move
\[ f^{-1}(x) = \sqrt[3]{\tfrac{x-1}{2}} \]
Verify: compose one way
Why: Substituting the inverse into f gives 2 times the cube of the cube root, plus 1, which is 2 times (x minus 1) over 2, plus 1, which is x. The operations undid each other in reverse order: the cube root undid the cube, the multiplication by 2 undid the division, and the plus 1 undid the minus 1.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 441-441
Ranking
Finding the inverse of a power function.
Put in order
Why: Steps three to five reverse the order in which the original function applied its operations, which is what solving the switched equation naturally produces. Step one comes first because it decides whether a restriction is needed before any algebra is attempted.
Worked example
Guided Practice 6 to 10.
\[ \text{Invert } g(x) = \tfrac{1}{27}x^3, \; f(x) = -\tfrac{64}{125}x^3, \; f(x) = 2x^5+3, \; g(x) = -7x^5+7. \]
First: switch and isolate
Why: X equals one twenty-seventh y cubed, so y cubed is 27x.
Second: switch and isolate
Why: X equals negative sixty-four over 125 y cubed, so y cubed is negative 125 over 64 times x.
Third: switch and isolate
Why: X equals 2y to the fifth plus 3, so y to the fifth is x minus 3, over 2.
Fourth: switch and isolate
Why: X equals negative 7y to the fifth plus 7, so y to the fifth is 7 minus x, over 7.
Figure (svg): The solution to Worked example four more power inverses shown as a ladder of expressions, one row per algebraic move
\[ 3\sqrt[3]{x}, \; -\tfrac{5}{4}\sqrt[3]{x}, \; \sqrt[5]{\tfrac{x-3}{2}}, \; \sqrt[5]{\tfrac{7-x}{7}} \]
Verify: check the second at a value
Why: G of 5 is negative 64 over 125 times 125, which is negative 64. The inverse of negative 64 should be 5: negative five quarters times the cube root of negative 64 is negative five quarters times negative 4, which is 5. The negative coefficient survived into the inverse, as it must for the composition to cancel.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 441-441
Error analysis
A student inverts a cubic.
Annotate
On: \( y = 2x^3+1 \;\Longrightarrow\; x = 2y^3+1 \;\Longrightarrow\; y = \sqrt[3]{\tfrac{x}{2}} - 1 \)
An inverse undoes operations in reverse order. Solving the switched equation step by step enforces that automatically, which is why the algebra is safer than reasoning it out in words.
Fill the middle
Example 5, at the third step.
Fill in the blanks
x = 2y^3+1 \;\Longrightarrow\; x - 1 = 2y^3 \;\Longrightarrow\; y^3 = \frac2___}
Why: Dividing both sides by 2 isolates y cubed, and the cube root then finishes it. The subtraction had to come before the division because in the original the multiplication came before the addition — the reverse order throughout.
Matching
Undo the operations backwards.
Match the pairs
Why: Every one has an odd power, so none needs a restriction and none has a plus-or-minus. The second is the tidiest: with no added constant the inverse is a clean multiple of a cube root, because the reciprocal of one twenty-seventh is 27 and the cube root of 27 is 3.
Prediction
Commit before reasoning.
Predict first
Why does an odd-power function always pass the horizontal line test?
Correct: Because it increases throughout, so it never repeats a value.
\[ a < b \;\Longrightarrow\; a^3 < b^3, \quad \text{but} \quad (-2)^2 = 2^2 \]
Why: An odd power preserves order: if a is less than b then a cubed is less than b cubed. So the graph rises from left to right without ever turning, and no horizontal line can meet it twice. An even power folds the negatives onto the positives, taking the same value at x and at negative x, which is precisely what makes it fail.
Section
Section 5
Concept
To invert a model, solve its equation for the other variable. The letters are not switched, because they were chosen to name real quantities and switching them would destroy their meaning.
\[ R = \tfrac{3}{8}L - 5 \;\Longrightarrow\; L = \tfrac{8}{3}R + \tfrac{40}{3} \]
To undo a power such as t to the 0.192, raise both sides to the reciprocal exponent. That is the power of a power rule from Lesson 6.2, with the two exponents multiplying to 1.
Figure (svg): Two models inverted, one linear and one a power function
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 439-442 — Examples 3 and 6
Picture it
Examples 3 and 6.
Figure (svg): Two models inverted, one linear and one a power function
One is undone by ordinary algebra and the other by a reciprocal exponent, but both answer the same kind of question: given the output, what was the input?
Worked example
Example 3 and Guided Practice 4.
\[ \text{With } R = \tfrac{3}{8}L - 5, \text{ find } L \text{ as a function of } R \text{ and evaluate at } R = 19 \text{ and } R = 13. \]
Undo the subtraction
Why: Adding 5 to both sides isolates the term in L.
\[ R + 5 = (\frac{3}{8}) L \]
Undo the coefficient
Why: Multiplying both sides by eight thirds.
\[ L = (\frac{8}{3}) R + \frac{40}{3} \]
Evaluate at 19 pounds
Why: Eight thirds of 19 is 152 over 3, plus 40 over 3 is 192 over 3.
\[ L = 64\text{ inches} \]
Evaluate at 13 pounds
Why: Eight thirds of 13 is 104 over 3, plus 40 over 3 is 144 over 3.
\[ L = 48\text{ inches} \]
Figure (svg): The solution to Worked example the elastic band shown as a ladder of expressions, one row per algebraic move
\[ L = \tfrac{8}{3}R + \tfrac{40}{3}; \quad 64 \text{ in}, \; 48 \text{ in} \]
Verify: substitute back into the original
Why: At L equal to 64, three eighths of 64 is 24, minus 5 is 19 pounds. At L equal to 48, three eighths of 48 is 18, minus 5 is 13 pounds. Both check. Note that the letters stayed as R and L throughout, so each equation still reads as a statement about resistance and length.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 439-439
Fill the middle
Example 6, at the second step.
Fill in the blanks
P/35 = t^5.2 \;\Longrightarrow\; t = (P/35)^___ \approx (P/35)^___}
Why: One over 0.192 is about 5.208, rounded to 5.2. Raising to the reciprocal exponent undoes a power because the two exponents multiply to 1 — the power of a power rule from Lesson 6.2 used deliberately rather than incidentally.
Worked example
Example 6. Undoing a power means a reciprocal exponent.
\[ \text{With } P = 35t^{0.192}, \text{ find } t \text{ as a function of } P. \]
Undo the coefficient
Why: Dividing both sides by 35 isolates the power.
\[ \frac{P}{35} = t ^{0.192} \]
Undo the power
Why: Raise both sides to the reciprocal exponent, 1 over 0.192.
\[ (\frac{P}{35}) ^{\frac{1}{0.192}} \]
Simplify the exponent
Why: One over 0.192 is about 5.2.
\[ (\frac{P}{35}) ^{5.2} \]
State the inverse
Why: The model now gives the year from the price.
\[ t = (\frac{P}{35}) ^{5.2} \]
Figure (svg): The solution to Worked example the ticket price model shown as a ladder of expressions, one row per algebraic move
\[ t \approx \left(\tfrac{P}{35}\right)^{5.2} \]
Verify: check that the exponents cancel
Why: Raising t to the 0.192 and then to the 1 over 0.192 multiplies the exponents to give t to the first, which is t. That is the power of a power rule doing exactly what it was needed for, and it is why the reciprocal exponent is the right way to undo a power.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 442-442
Trap
\[ R = \tfrac{3}{8}L - 5 \]
Switch R and L as for an abstract relation
Why: The usual x and y swap is applied to the model's letters.
\[ L = \tfrac{3}{8}R - 5 \quad \text{(wrong, and now meaningless)} \]
This says the length equals three eighths of the resistance minus 5, which is not what the model claims and does not agree with it at any value.
\[ R = \tfrac{3}{8}L - 5 \;\Longrightarrow\; L = \tfrac{8}{3}R + \tfrac{40}{3} \]
Solve for the other letter, leaving both names alone
Why: R still means resistance and L still means length; only which one is the subject has changed.
\[ R = 19 \;\Longrightarrow\; L = 64 \]
The textbook makes this point explicitly in a Find Inverses note. Switching letters is a device for abstract relations in x and y; with named quantities it destroys the meaning.
Sorting
Ask whether the letters name real quantities.
Sort into buckets
Sort each situation.
The distinction is about meaning rather than mathematics: both procedures produce the same relationship, but only one keeps the equation readable.
Prediction
Commit before reasoning.
Predict first
The original model gives the ticket price from the year. What does its inverse give?
Correct: The year from the price.
\[ P = 35t^{0.192} \text{ gives } P; \quad t = (P/35)^{5.2} \text{ gives } t \]
Why: Inverting swaps which quantity is known and which is being found. That is exactly what the elastic band example needed too: the original gave resistance from length, and the question asked for the length that produces a given resistance. Inverting a model is the standard response to a question that runs the wrong way through it.
Comparison
Fill the blanks. Same idea, different bookkeeping.
Comparison matrix
| Question | Abstract relation | Model |
|---|---|---|
| First move | switch x and y | solve for the other letter |
| Why | the letters are placeholders | the letters name quantities |
| Result reads as | a new function of x | the same relationship, rearranged |
| Undoing a power | take the matching root | raise to the reciprocal exponent |
The last row is two descriptions of one operation: the nth root and the one-over-n power are the same thing, as Lesson 6.1 established.
Comparison
Fill the blanks. Lesson 6.3 set this up; this lesson uses it.
Comparison matrix
| Question | General functions, 6.3 | Inverse functions, 6.4 |
|---|---|---|
| Does order matter? | yes, almost always | no: both orders give x |
| The result of composing | usually a new, unrelated function | the identity: the input itself |
| How you check | substitute and simplify | substitute and check both give x |
| Graphical relationship | none in particular | reflections in the line y = x |
Inverse functions are the special case that makes composition worth defining carefully, which is why the two lessons are adjacent.
Pattern
One routine for finding an inverse and confirming it.
The notation f to the negative 1 names the inverse function. It never means one over f, and reading it that way produces nonsense.
OpenStax Algebra and Trigonometry 2e, §3.7 Inverse Functions §3.7
Check
Finding an inverse. Switch, then solve.
Check your understanding
What is the inverse of f(x) = 2x - 1?
Answer: A
Why: Switching gives x = 2y - 1, so 2y = x + 1 and y = (x + 1)/2.
Check
The horizontal line test.
Check your understanding
For which of these is the inverse a function without any domain restriction?
Answer: A
Why: An odd power increases throughout, so no horizontal line meets its graph twice.
Check
Inverting a cubic. Undo in reverse order.
Check your understanding
What is the inverse of f(x) = 2x^3 + 1?
Answer: A
Why: Switching gives x = 2y^3 + 1, so y^3 = (x - 1)/2 and y is its cube root.
Real world
A recipe converts oven temperature between scales with F equals 1.8C plus 32. A separate rule gives the surface area of a sphere as S equals 4 pi r squared.
Discussion prompt
Invert both, say which one needs a domain restriction and why, and use the second to find the radius of a ball with surface area 200 square centimetres.
Hint: One is linear and one is an even power.
Answer:
\[ F = 1.8C + 32 \;\Longrightarrow\; C = \frac{F-32}{1.8} \]
\[ S = 4\pi r^2 \;\Longrightarrow\; r = \sqrt{\frac{S}{4\pi}} \quad (r \ge 0) \]
The temperature conversion needs no restriction: a line passes the horizontal line test, so its inverse is a function everywhere. The sphere formula does need one — but the restriction is free, because a radius cannot be negative in the first place.
\[ r = \sqrt{\frac{200}{4\pi}} = \sqrt{15.92} \approx 3.99 \text{ cm} \]
Two things are worth noticing. The situation supplied the restriction that the mathematics required, which is common in models and is why the plus-or-minus so rarely causes trouble there. And neither inversion switched the letters: C still means Celsius and r still means radius, exactly as the textbook's note insists.
Commit first
Answer, then rate your confidence honestly.
Predict first
Does f to the negative 1 of x mean one over f of x?
Correct: No — it names the inverse function.
\[ f^{-1}(x) = \tfrac{1}{3}x + \tfrac{5}{3} \quad \text{but} \quad \tfrac{1}{f(x)} = \tfrac{1}{3x-5} \]
Why: For f of x equal to 3x minus 5, the inverse is one third x plus five thirds while one over f of x is 1 over 3x minus 5. At x equal to 4 the inverse gives 3 and the reciprocal gives one seventh. The superscript is part of a name, not an operation, and the textbook flags the clash in a Reading note. The two coincide only in rare accidents, never as a rule.
Explain it
They can solve equations and have just seen the inverse notation for the first time.
Discussion prompt
In four sentences or fewer, explain what an inverse function is and how you find one, without using the word inverse more than once.
Hint: Talk about undoing.
Answer:
A function turns an input into an output by doing a sequence of operations; its inverse is the function that turns that output back into the original input by undoing each operation in reverse order. To find it, write y for the output, swap x and y, and solve the result for y.
You can check you have it right by feeding a number through one function and then the other: you should end up where you started, and that has to work in both orders. Graphically, the two functions are mirror images across the line y equals x.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For solving, undo the operations in reverse order and write each step separately. For restrictions, apply the horizontal line test before any algebra. For signs, match the inverse's range to the original's domain. For letters, ask whether they name real quantities — if they do, solve rather than switch. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take f of x equals 2x cubed plus 1 and work it through completely. Top left: sketch it, draw the line y equals x, and sketch the reflection, marking two pairs of points whose coordinates are reverses of one another. Top right: find the inverse algebraically, one line per operation undone, and note beside each line which operation of the original it is reversing. Bottom left: verify both compositions give x, showing every cancellation. Bottom right: repeat the whole exercise for f of x equals x squared, and write one sentence saying what has to be added before the inverse is a function and why. In a margin, write what f to the negative 1 does not mean.
If your two sketches look like reflections of each other but your algebra disagrees, trust the algebra and recheck the order in which you undid the operations.
Recap
Five things, and the second is the definition the other four serve.
| If you see | Then |
|---|---|
| An abstract relation in x and y | Switch the variables and solve |
| A model with named quantities | Solve for the other letter |
| An even power | Restrict the domain first |
| An odd power | No restriction is needed |
| A claimed pair of inverses | Check both compositions |
| f to the negative 1 | Read it as a name, not a reciprocal |
| A power to undo in a model | Raise to the reciprocal exponent |
Lesson 6.5 graphs the functions this lesson produced: the square root and cube root functions, which are the inverses of the squaring and cubing functions.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.4 Use Inverse Functions §6.4, pp. 438-445 — everything on these slides traces back here
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