Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.3 - New Functions from Old Functions - 1.3 Exercises - Page 44: 35


(a) $$f(g(x))=\sqrt{4x-2}$$ $$[\frac{1}{2}, \infty)$$ (b) $$g(f(x))= 4\sqrt{x+1}-3$$ $$[-1, \infty)$$ (c) $$f(f(x))= \sqrt{\sqrt{x+1}+1}$$ $$[-1, \infty)$$ (d) $$g(g(x))= 16x-15$$ $$(-\infty, \infty)$$

Work Step by Step

$f(x)=\sqrt{x+1}$ $g(x)=4x-3$ (a)$ f(g(x))= \sqrt{4x-3+1} = \sqrt{4x-2}$ Since we have a square root in the function and we should have a non-negative number under the square root, the domain will be: $4x-2\geq0$ $4x\geq2$ $x\geq \frac{1}{2}$ $[\frac{1}{2}, \infty)$ (b) $g(f(x))= 4\sqrt{x+1}-3$ We have the same restriction here: $x+1\geq0$ $x\geq-1$ $[-1, \infty)$ (c) $f(f(x))= \sqrt{\sqrt{x+1}+1}$ This time we have something like a "double restriction": $Domain= \left\{ \begin{array}{ll} x+1\geq0 \\ \sqrt{x+1}+1\geq0 \end{array} \right. \left\{ \begin{array}{ll} x\geq-1 \\ \sqrt{x+1}\geq -1 \end{array} \right.$ The square root of a number is always greater than a negative number, so we can omit the second expression and say that the domain of this function is: $[-1, \infty)$ (d) $g(g(x))= 4(4x-3)-3 = 16x-12-3 = 16x-15$ We have no restriction in this case, so the domain is: $(-\infty, \infty)$
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