Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.1 Review of Functions - 1.1 Exercises: 81

Answer

$a.$ True. $b.$ Not true. $c.$ True. $d.$ Not true. $e.$ Not true. $f.$ True. $g.$ True. $h.$ Not true. $i.$ True.

Work Step by Step

$a.$ True. $\forall y\in\mathbb{R},\exists x.y=2x-38$. $b.$ Not true. In the definition of a function, it is allowed for different $x$-es to map to the same $y$. $c.$ True. $f(1/x)=(1/x)^{-1}=x=\frac{1}{x^-1}=\frac{1}{f(x)}.$ $d.$ Not true. The counter example is $f(x)=2x$. Then $f(f(x))=f(2x)=2(2x)=4x\neq (f(x))^2.$ $e.$ Not true. The counter example is $f(x)=\sin x$ and $g(x)=2x.$ Then we have $f(g(x))=\sin(g(x))=\sin(2x)$ and $g(f(x))=2f(x)=2\sin x$. Obviously $\sin(2x)\neq2\sin x$. $f.$ True. $(f\circ g)(x)$ is just another way to write $f(g(x))$. $g.$ True. If $f$ is an even function then $f(-x)=f(x)$. Then also $cf(a\cdot(-x))=cf(-ax)=cf(ax)$. $h.$ Not true. If $f$ is an odd function then $f(-x)=-f(x)$. Then we have $f(-x)+d=-f(x)+d\neq-(f(x)+d)$. $i.$ True. If $f(x)$ is both even and odd then for all $x$ $f(-x)=f(x)$ and $f(-x)=-f(x)$ which gives $f(x)=-f(x)\Rightarrow f(x)=0$.
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