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 - Chapter Test - Page 427: 23

Answer

$(g \circ f)(x)$ = $g(f(x)) = 4x^2 - 7$ $(f \circ g)(x) = f(g(x)) = 16x^2 + 8x - 1$

Work Step by Step

For these types of problems, we are asked to evaluate composite functions. We will use the inner function and substitute it where we see $x$ in the outer function. We can rewrite $(g \circ f)(x)$ as $g(f(x))$. We now begin by plugging in the inner function $f(x)$ where we see $x$ in the outer function, $g(x)$: $(g \circ f)(x)$ = $g(f(x)) = 4(x^2 - 2) + 1$ Use distributive property to get rid of the parentheses: $(g \circ f)(x)$ = $g(f(x)) = 4(x^2) - 4(2) + 1$ Multiply to simplify: $(g \circ f)(x)$ = $g(f(x)) = 4x^2 - 8 + 1$ Combine like terms: $(g \circ f)(x)$ = $g(f(x)) = 4x^2 - 7$ We can rewrite $(f \circ g)(x)$ as $f(g(x))$. We now begin by plugging in the inner function $g(x)$ where we see $x$ in the outer function, $f(x)$: $(f \circ g)(x) = f(g(x)) = (4x + 1)^2 - 2$ Let's rewrite to expand the binomial: $(f \circ g)(x) = f(g(x)) = (4x + 1)(4x + 1) - 2$ We're going to use the FOIL method to multiply the two binomials. With the FOIL method, we multiply the first terms first, then the outer terms, then the inner terms, and, finally, the last terms: Evaluate the exponent: $(f \circ g)(x) = f(g(x)) = [(4x)(4x) + (4x)(1) + (1)(4x) + (1)(1)] - 2$ Let's multiply out the terms: $(f \circ g)(x) = f(g(x)) = (16x^2 + 4x + 4x + 1) - 2$ Combine like terms: $(f \circ g)(x) = f(g(x)) = (16x^2 + 8x + 1) - 2$ Subtract: $(f \circ g)(x) = f(g(x)) = 16x^2 + 8x - 1$
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