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Algebra 2 Honors · Lesson 1.3

Composition and Inverses

How it works

Composition applies one function and then the other, so order matters. The output of the inner function must lie in the outer function domain. An inverse reverses a one-to-one relationship; finding a formula involves swapping input and output roles and solving.

(f∘g)(x)=f(g(x)),f−1(f(x))=x(f\circ g)(x)=f(g(x)),\qquad f^{-1}(f(x))=x

Worked examples

Example 1

For f(x)=2x+1\displaystyle f(x)=2x+1, g(x)=x2\displaystyle g(x)=x^2, find f(g(3))\displaystyle f(g(3)).

g(3)=9\displaystyle g(3)=9, then f(9)=19\displaystyle f(9)=19. Reversing the order gives a different expression.

Example 2

Find the inverse of f(x)=3x−4\displaystyle f(x)=3x-4.

Write y=3x−4\displaystyle y=3x-4, swap roles, and solve x=3y−4\displaystyle x=3y-4. Thus f−1(x)=(x+4)/3\displaystyle f^{-1}(x)=(x+4)/3.

Problems

Problem 1.3.1

For f(x)=x+2\displaystyle f(x)=x+2 and g(x)=3x\displaystyle g(x)=3x, find (g∘f)(x)\displaystyle (g\circ f)(x).

Show worked solution

g(f(x))=3(x+2)=3x+6\displaystyle g(f(x))=3(x+2)=3x+6.

Problem 1.3.2

Find the inverse of h(x)=(x−5)/2\displaystyle h(x)=(x-5)/2.

Show worked solution

From y=(x−5)/2\displaystyle y=(x-5)/2, solve for x=2y+5\displaystyle x=2y+5. Thus h−1(x)=2x+5\displaystyle h^{-1}(x)=2x+5.

Problem 1.3.3

The graph shows y=x2\displaystyle y=x^2 restricted to x≥0\displaystyle x\ge0 and its inverse. State the inverse and explain the restriction.

The right half of y = x squared and y = square root of x reflect across y = x.
The restricted parabola passes the horizontal-line test.
Show worked solution

The inverse is y=x\displaystyle y=\sqrt{x}, with domain and range [0,∞)\displaystyle [0,\infty). Restricting the parabola to x≥0\displaystyle x\ge0 makes it one-to-one; the graphs reflect across y=x\displaystyle y=x.

The right half of y = x squared and y = square root of x reflect across y = x. The inverse is square root of x; both domains are nonnegative.
The inverse is square root of x; both domains are nonnegative.

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