Mathfolis

Inverse Functions

Unit 2 · Exponential and Logarithmic Functions

What AP Precalc asks here

An inverse function f1f^{-1} undoes what ff does: f(f1(x))=xf(f^{-1}(x)) = x and f1(f(x))=xf^{-1}(f(x)) = x. The inverse exists if and only if ff is one-to-one (passes the horizontal line test). Algebraically: write y=f(x)y = f(x), swap xx and yy, solve for yy. Graphically, the inverse is the reflection across y=xy = x; domain and range swap.

Inverse identities

Composition
f(f1(x))=x   and   f1(f(x))=xf(f^{-1}(x)) = x \;\text{ and }\; f^{-1}(f(x)) = x
Reflection
graph of f1=reflection of f across y=x\text{graph of } f^{-1} = \text{reflection of } f \text{ across } y = x
Domain/range swap
dom(f)=range(f1),  range(f)=dom(f1)\text{dom}(f) = \text{range}(f^{-1}), \; \text{range}(f) = \text{dom}(f^{-1})
Caution: f1(x)f^{-1}(x) is the inverse function, not the reciprocal 1/f(x)1/f(x). The exponent 1-1 in inverse notation is just notation.
Type 1

Find the inverse algebraically

Write y=f(x)y = f(x), swap xx and yy, then solve for the new yy.

Example 1
Find f1(x)f^{-1}(x) for f(x)=3x7f(x) = 3x - 7. (A) f1(x)=x+73f^{-1}(x) = \dfrac{x + 7}{3} (B) f1(x)=x73f^{-1}(x) = \dfrac{x - 7}{3} (C) f1(x)=3x+7f^{-1}(x) = 3x + 7 (D) f1(x)=13x7f^{-1}(x) = \dfrac{1}{3x - 7}

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Type 2

Verify two functions are inverses

Compute f(g(x))f(g(x)) and g(f(x))g(f(x)). Both must simplify to xx.

Example 2
Are f(x)=2x+4f(x) = 2x + 4 and g(x)=x42g(x) = \dfrac{x - 4}{2} inverses? (A) Yes — both compositions simplify to xx (B) No — f(g(x))xf(g(x)) \neq x (C) No — g(f(x))xg(f(x)) \neq x (D) Cannot be determined without a graph

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Type 3

Use the graph/point/domain swap

Swap coordinates of points; swap domain and range; reflect graphs across y=xy = x.

Example 3
If the point (3,7)(3, 7) is on the graph of ff, what is the corresponding point on the graph of f1f^{-1}? (A) (3,7)(-3, -7) (B) (3,7)(3, 7) (C) (7,3)(7, 3) (D) (1/3,1/7)(1/3, 1/7)

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