Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 3 - Section 3.1 - Quadratic Functions and Models - 3.1 Exercises - Page 252: 49

Answer

$7$

Work Step by Step

Given: $f(x)=3+4x^2-x^4$ Let us consider $x=t, t=x^2, t^2=x^4$ Now, we have $f(t)=-t^2+4t+3$ Here, $a=-1, b=4, c=3$ But $a=-1 \lt 0$ shows that the function has a maximum value. The maximum value occurs at $t=-\dfrac{b}{2a}=-\dfrac{4}{2(-1)}=2$ Thus, the maximum value of the function $f(x)=3+4x^2-x^4$ occurs is $f(-\dfrac{b}{2a})$. $f(2)=3+4(2)-(2)^2$ This gives: $f(2)=7$
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