Section Navigation

Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 6 - Radical Functions and Rational Exponents - 6-7 Inverse Relations and Functions - Practice and Problem-Solving Exercises - Page 412: 84

Answer

$(f \circ g)(x) = f(g(x)) = 2x + 28$

Work Step by Step

In this problem, we are asked to evaluate a composite function. We will use the inner function and substitute it where we see $x$ in the outer function. This means that $(f \circ g)(x) = f(g(x))$, so where we see $x$ in the $f(x)$ function, we will plug in the function $g(x)$. To evaluate $f(g(x))$, we begin by plugging in the function $g(x)$ where we see $x$ in the outer function, $f(x)$: $f(g(x)) = 4(\frac{1}{2}x + 7)$ Use distributive property to multiply the coefficient with each term of the binomial in the composite function first: $f(g(x)) = (4)(\frac{1}{2}x) + (4)(7)$ Multiply to simplify: $f(g(x)) = \frac{4}{2}x + 28$ Divide numerator and denominator by their greatest common factor, which is $2$, in this case: $f(g(x)) = 2x + 28$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.