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Logarithmic, Exponential, and Other Transcendental Functions

5.3 Inverse Functions

Focus on recover an input, inverse derivative, composition with an inverse.

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Ideas in this lesson

01Recover an inputWorked steps · five questions02Inverse derivativeWorked steps · five questions03Composition with an inverseWorked steps · five questions
Idea 1 of 3

Recover an input

An inverse undoes a one-to-one function. Interchanging coordinates is different from taking a reciprocal.

Worked example 1

Let f(x)=2x+1f(x) = 2x + 1. Find f−1(3)f^{-1}(3).
  1. y=2x+1  ⟹  x=y−12y = 2x + 1 \implies x = \dfrac{y - 1}{2}
  2. f−1(y)=y−12f^{-1}(y) = \dfrac{y - 1}{2}; (f−1)′(y)=12(f^{-1})'(y) = \dfrac{1}{2}
  3. f−1(3)=3−12=1f^{-1}(3) = \dfrac{3 - 1}{2} = 1

Answer: 11

Worked example 2

Let f(x)=3x+1f(x) = 3x + 1. Find (f−1)′(7)(f^{-1})'(7).
  1. y=3x+1  ⟹  x=y−13y = 3x + 1 \implies x = \dfrac{y - 1}{3}
  2. f−1(y)=y−13f^{-1}(y) = \dfrac{y - 1}{3}; (f−1)′(y)=13(f^{-1})'(y) = \dfrac{1}{3}
  3. (f−1)′(7)=13≈0.33(f^{-1})'(7) = \dfrac{1}{3} \approx 0.33

Answer: 13≈0.33\dfrac{1}{3} \approx 0.33

Idea 2 of 3

Inverse derivative

At corresponding inputs, inverse-function slopes are reciprocal when the original slope is nonzero.

Worked example 1

For f(x)=x3f(x) = x^{3}, find f−1(8)f^{-1}(8).
  1. f−1(y)=y3f^{-1}(y) = \sqrt[3]{y}; f′(2)=12f'(2) = 12
  2. (f−1)′(8)=112(f^{-1})'(8) = \dfrac{1}{12}
  3. 83=2\sqrt[3]{8} = 2

Answer: 22

Worked example 2

For f(x)=x3f(x) = x^{3}, find (f−1)′(27)(f^{-1})'(27).
  1. f−1(y)=y3f^{-1}(y) = \sqrt[3]{y}; f′(3)=27f'(3) = 27
  2. (f−1)′(27)=127(f^{-1})'(27) = \dfrac{1}{27}
  3. Inverse slope =127≈0.04= \dfrac{1}{27} \approx 0.04

Answer: 127≈0.04\dfrac{1}{27} \approx 0.04

Idea 3 of 3

Composition with an inverse

The identities f−1(f(x))=xf^{-1}(f(x)) = x and f(f−1(y))=yf(f^{-1}(y)) = y hold on their appropriate domains.

Worked example 1

Let f(x)=2x+1f(x) = 2x + 1. Find f−1(3)f^{-1}(3).
  1. y=2x+1  ⟹  x=y−12y = 2x + 1 \implies x = \dfrac{y - 1}{2}
  2. f−1(y)=y−12f^{-1}(y) = \dfrac{y - 1}{2}; (f−1)′(y)=12(f^{-1})'(y) = \dfrac{1}{2}
  3. f−1(3)=3−12=1f^{-1}(3) = \dfrac{3 - 1}{2} = 1

Answer: 11

Worked example 2

Let f(x)=3x+1f(x) = 3x + 1. Find (f−1)′(7)(f^{-1})'(7).
  1. y=3x+1  ⟹  x=y−13y = 3x + 1 \implies x = \dfrac{y - 1}{3}
  2. f−1(y)=y−13f^{-1}(y) = \dfrac{y - 1}{3}; (f−1)′(y)=13(f^{-1})'(y) = \dfrac{1}{3}
  3. (f−1)′(7)=13≈0.33(f^{-1})'(7) = \dfrac{1}{3} \approx 0.33

Answer: 13≈0.33\dfrac{1}{3} \approx 0.33

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §5.3,p5.3, p. 337337. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.