Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 5 - Graphs and the Derivative - 5.3 Higher Derivatives, Concavity, and the Second Derivative Test - 5.3 Exercises - Page 284: 35


$$ f(x)=-2x^{3}+9x^{2}+168x-3 $$ $f(x)$ concave upward on $$(-\infty,\frac{3}{2} )$$. $f(x)$ concave downward on $$(\frac{3}{2}, \infty )$$ Inflection point at $$(\frac{3}{2}, \frac{525}{2})$$

Work Step by Step

$$ f(x)=-2x^{3}+9x^{2}+168x-3 $$ The first derivative is $$ \begin{aligned} f^{\prime}(x) &=-2 (3)x^{2}+9(2)x+168\\ &=-6x^{2}+18x+168, \end{aligned} $$ and the second derivative is $$ \begin{aligned} f^{\prime\prime}(x) &=-6(2)x+18+0\\ &=-12x+18. \end{aligned} $$ Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$. $$ \begin{aligned} f^{\prime\prime}(x) =-12x+18 &=0\\ -12x &=-18\\ x &=\frac{-18}{-12}\\ x &=\frac{3}{2}. \end{aligned} $$ * Test a number in the interval $(-\infty,\frac{3}{2} )$ to see that $f^{\prime\prime}(x)$ is positive there, so $f(x)$ is concave upward on $(-\infty,\frac{3}{2} )$. ** Test a number in the interval $(\frac{3}{2}, \infty )$to find $f^{\prime\prime}(x)$ negative in that interval, so $f(x)$ is concave downward on $(\frac{3}{2}, \infty )$. *** Since the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $\frac{3}{2}$, the graph changes from concave upward to concave downward at that point, so $(\frac{3}{2}, f(\frac{3}{2}))$ is inflection point. Now ,we find $f(\frac{3}{2})$as follows: $$ f(\frac{3}{2})=-2(\frac{3}{2})^{3}+9(\frac{3}{2})^{2}+168(\frac{3}{2})-3=\frac{525}{2} $$ Finally, we have inflection point at $(\frac{3}{2}, f(\frac{3}{2}))=(\frac{3}{2}, \frac{525}{2})$.
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