Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 12 - Section 12.1 - The Algebra of Functions - Exercise Set: 12


$(f\circ h)(1)=-3$

Work Step by Step

$f(x)=x^{2}-6x+2;$ $h(x)=\sqrt{x}$ $(f\circ h)(1)$ First, find $h(1)$ by substituting $x$ by $1$ in $h(x)$: $h(1)=\sqrt{1}=1$ Find $(f\circ h)(1)$ by substituting $x$ by $1$ in $f(x)$: $(f\circ h)(1)=f(1)=1^{2}-6(1)+2=1-6+2=-3$
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