Skills Beta

Inverse Functions

An inverse function runs a one-to-one function backward: f⁻¹(b) = a exactly when f(a) = b .

Practice 2: Connecting Representations

Question set for this topic

Part 1 · Hook

Why this matters

A function turns inputs into outputs. Often you know the output and need the input: what time gave this temperature, what angle has this sine. That is the job of an inverse function. In Unit 3 you will find how fast an inverse changes, using one fact you can see in a picture: reflecting a graph across y=xy = x flips its slopes.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If f(x)=x2f(x) = x^2 and g(x)=x+1g(x) = x + 1, what is f(g(x))f(g(x))?

  1. (x+1)2(x + 1)^2
  2. x2+1x^2 + 1
  3. x2(x+1)x^2(x + 1)
  4. 2x+12x + 1
Show the answer

g first, then f.

  • Correct: (x+1)2(x + 1)^2:
  • x2+1x^2 + 1:
  • x2(x+1)x^2(x + 1):
  • 2x+12x + 1:

2. What is sin⁡π6\sin\frac{\pi}{6}?

  1. 12\frac{1}{2}
  2. 32\frac{\sqrt{3}}{2}
  3. 22\frac{\sqrt{2}}{2}
  4. 11
Show the answer

A first-quadrant special value.

  • Correct: 12\frac{1}{2}:
  • 32\frac{\sqrt{3}}{2}:
  • 22\frac{\sqrt{2}}{2}:
  • 11:

Part 4 · See it

See it first

The curve y = f(x), with f(x) = 0.5x cubed + 0.5x + 2, passes through (1, 3); its inverse, the mirror image across the dashed line y = x, passes through (3, 1). Short lines show slope 2 on f at (1, 3) and slope 1/2 on the inverse at (3, 1).
ff and f−1f^{-1} are mirror images across y=xy = x. The point (1,3)(1, 3) on ff becomes (3,1)(3, 1) on f−1f^{-1}, and the slope 2 there becomes 12\frac{1}{2}. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. ff sends each input aa to an output bb, and no two inputs share an output (one-to-one)the inverse function f−1f^{-1} sends bb back to aa: f−1(b)=af^{-1}(b) = a exactly when f(a)=bf(a) = b
  2. Swapping inputs and outputsswaps the coordinates of every point, reflecting the graph across y=xy = x
  3. Swapping coordinates swaps rise and runslopes at matching points are reciprocals
  4. Sine, cosine and tangent repeat, so they are not one-to-onetheir inverses use restricted ranges: arcsine and arctangent give angles from −π/2 to π/2, arccosine from 0 to π

Part 6 · Key ideas

Key ideas

  • f−1(b)=af^{-1}(b) = a means f(a)=bf(a) = b. So f(f−1(x))=xf(f^{-1}(x)) = x and f−1(f(x))=xf^{-1}(f(x)) = x on the right domains.
  • f−1(x)f^{-1}(x) is not 1f(x)\frac{1}{f(x)}. The −1-1 means "undo", not "reciprocal".
  • Only a one-to-one function (each output from one input; passes the horizontal line test) has an inverse. Restrict the domain if needed.
  • To find a formula: write x=f(y)x = f(y) and solve for yy.
  • arcsin⁡x\arcsin x and arctan⁡x\arctan x return angles in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]; arccos⁡x\arccos x returns angles in [0,π][0, \pi].

Part 7 · Misconception

A common mistake

The wrong idea: f−1(x)=1f(x)f^{-1}(x) = \frac{1}{f(x)}, and sin−1⁡x=1sin⁡x\sin^{-1} x = \frac{1}{\sin x}.

What actually happens: f−1f^{-1} undoes ff: if f(2)=7f(2) = 7 then f−1(7)=2f^{-1}(7) = 2. The reciprocal 1f(x)\frac{1}{f(x)} is a different function. sin−1⁡x\sin^{-1} x is arcsin⁡x\arcsin x; 1sin⁡x\frac{1}{\sin x} is csc⁡x\csc x.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

An increasing function ff

Selected values of an increasing function ff; every output of ff comes from exactly one input
xx01234
f(x)f(x)356911

1. Find f−1(9)f^{-1}(9).

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

f(3)=9f(3) = 9, so f−1(9)=3f^{-1}(9) = 3. Read the table from the output row back to the input row.

  • Answer: 3

2. The average rate of change of ff on [2,3][2, 3] is 3. Find the average rate of change of f−1f^{-1} on [6,9][6, 9].

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

f−1(6)=2f^{-1}(6) = 2 and f−1(9)=3f^{-1}(9) = 3, so the rate is 3−29−6=13\frac{3 - 2}{9 - 6} = \frac{1}{3}, the reciprocal of ff's rate of 3 on [2,3][2, 3].

  • Answer: 13\frac{1}{3} ≈ 0.333

3. Which statement about ff is true?

  1. f−1(3)=0f^{-1}(3) = 0
  2. f−1(3)=1f(3)=19f^{-1}(3) = \frac{1}{f(3)} = \frac{1}{9}
  3. f−1(3)=f(3)=9f^{-1}(3) = f(3) = 9
  4. f−1(3)f^{-1}(3) is not defined.
Show the answer

f(0)=3f(0) = 3, so f−1(3)=0f^{-1}(3) = 0. The superscript −1-1 means "inverse", not "reciprocal".

  • Correct: f−1(3)=0f^{-1}(3) = 0: f(0)=3f(0) = 3 read backward.
  • f−1(3)=1f(3)=19f^{-1}(3) = \frac{1}{f(3)} = \frac{1}{9}: f−1f^{-1} is the inverse function; 1f(3)\frac{1}{f(3)} is a reciprocal, a different thing.
  • f−1(3)=f(3)=9f^{-1}(3) = f(3) = 9: That reads the table forward. The inverse goes from output to input.
  • f−1(3)f^{-1}(3) is not defined.: 3 is an output of ff (at x=0x = 0), so f−1(3)f^{-1}(3) exists.

4. Let f(x)=2x+1x−3f(x) = \frac{2x + 1}{x - 3}. Find f−1(x)f^{-1}(x).

Type an expression for f−1(x)=f^{-1}(x) =.

Show the answer

Write x=2y+1y−3x = \frac{2y + 1}{y - 3} and solve for yy: xy−3x=2y+1xy - 3x = 2y + 1, so xy−2y=3x+1xy - 2y = 3x + 1, so y(x−2)=3x+1y(x - 2) = 3x + 1 and y=3x+1x−2y = \frac{3x + 1}{x - 2}. Check: f(0)=−13f(0) = -\frac{1}{3} and f−1(−13)=0−7/3=0f^{-1}\left(-\frac{1}{3}\right) = \frac{0}{-7/3} = 0.

  • Answer: f−1(x)=f^{-1}(x) = 3x+1x−2\frac{3x + 1}{x - 2}

5. Find arccos⁡(−12)\arccos\left(-\frac{1}{2}\right) exactly.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Arccosine gives an angle in [0,π][0, \pi]. cos⁡2π3=−12\cos\frac{2\pi}{3} = -\frac{1}{2}, so arccos⁡(−12)=2π3\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}.

  • Answer: 2π3\frac{2\,\pi}{3} ≈ 2.094

6. Find arcsin⁡(sin⁡5π6)\arcsin\left(\sin\frac{5\pi}{6}\right) exactly.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Inside first: sin⁡5π6=12\sin\frac{5\pi}{6} = \frac{1}{2}. Then arcsin⁡12=π6\arcsin\frac{1}{2} = \frac{\pi}{6}. The two functions do not cancel here, because 5π6\frac{5\pi}{6} is outside arcsine's range.

  • Answer: π6\frac{\pi}{6} ≈ 0.524

7. The point (2,7)(2, 7) is on the graph of a one-to-one function ff, and the slope of the graph of ff there is 4. Which is true about the graph of f−1f^{-1}?

  1. It passes through (7,2)(7, 2), with slope 14\frac{1}{4} there.
  2. It passes through (7,2)(7, 2), with slope 4 there.
  3. It passes through (2,7)(2, 7), with slope 14\frac{1}{4} there.
  4. It passes through (−2,−7)(-2, -7), with slope −4-4 there.
Show the answer

Reflecting across y=xy = x swaps coordinates, so (2,7)(2, 7) becomes (7,2)(7, 2), and swaps rise and run, so slope 4 becomes 14\frac{1}{4}.

  • Correct: It passes through (7,2)(7, 2), with slope 14\frac{1}{4} there.: Swap the point and flip the slope.
  • It passes through (7,2)(7, 2), with slope 4 there.: The reflection swaps rise and run, so the slope flips to its reciprocal.
  • It passes through (2,7)(2, 7), with slope 14\frac{1}{4} there.: The point must be swapped too.
  • It passes through (−2,−7)(-2, -7), with slope −4-4 there.: Reflection across y=xy = x swaps coordinates; it does not negate them.

Part 9 · Summary

Summary

An inverse function runs a one-to-one function backward: f−1(b)=af^{-1}(b) = a exactly when f(a)=bf(a) = b. Its graph is the reflection of ff across y=xy = x, so points swap coordinates and slopes become reciprocals. Find a formula by solving x=f(y)x = f(y) for yy. The inverse trig functions use restricted ranges, and f−1f^{-1} never means 1f\frac{1}{f}.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections