College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 3, Polynomial and Rational Functions - Section 3.1 - Quadratic Functions and Models - 3.1 Exercises - Page 288: 50

Answer

The minimum value is $-14$ attained when $x=-\sqrt[3] 2$.

Work Step by Step

As suggested, we use the substitution $t=x^3$ to get $f(x) = 2 + 16t+4t^2$ which is a quadratic function of the form $f(t) = at^2+bt+c$. To find the minimum value, we find the vertex using the methods outlined in the chapter; mainly, $h=-\frac{b}{2a}$ and $k=f(h)$ with the vertex being the point $(h, k).$ We plug the values to get $h=-2$ and $k=-14.$ We note that this would represent the set of values of $f(t)$ rather than the required $f(x)$. To get the values for $f(x)$, we plug in $t=x^3$ to get that $x=\sqrt[3]{t}$ and hence the minimum value of $f(x)$ is $-14$ attained when $x=-\sqrt[3]{2}$. Attached below is a graph illustrating the curves mentioned above and the respective minimum points.
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