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Find the second derivative of f(x) = x^{3} + x^{4} and its sign chart. Which of the following is...

Question:

Find the second derivative of {eq}\displaystyle\;f(x) = x^{3} + x^{4}\; {/eq} and its sign chart.

Which of the following is not true about {eq}f {/eq}?

(a) {eq}\;f \textrm{ is concave down on } \left(-\frac{1}{2},0\right) {/eq}

(b) {eq}\;f \textrm{ is concave up on } \left(-\infty, -\frac{1}{2}\right) \cup \left(0,\infty\right) {/eq}

(c) {eq}\;f \textrm{ is concave up on } \left(-\frac{1}{2}, 0\right) {/eq}

(d) {eq}\;f \textrm{ has inflection points at } x = -\frac{1}{2} \textrm{ and } x = 0. {/eq}

(e) {eq}\textrm{ All are not true.} {/eq}

Concavity and Inflection points:

A graph is called concave if the tangent line of the graph at this point is under the graph near the point and concave down when the tangent line lies above the graph near the point. A point where the concavity moves (from the top to the bottom to the top) is called an inflection point.

Answer and Explanation:

Given that {eq}\displaystyle\;f(x) = x^{3} + x^{4}\; {/eq}

Differentiate twice with respect to x, we get

{eq}f'(x)=3x^2+4x^3\\ f''(x)=6x+12x^2\\ f''(x)=0\\ 6x+12x^2=0\\ 6x(1+2x)=0\\ x=0,\frac{-1}{2}\\ \mathrm{Sign\:of\:}f\:''\left(x\right)\mathrm{\:at\:}-\infty \:<x<\frac{-1}{2} \, is \quad \mathrm{Positive}\\ \therefore f(x) \, \textrm{is Concave up}\\ \mathrm{Sign\:of\:}f\:''\left(x\right)\mathrm{\:at\:}\frac{-1}{2}<x<0 \, is \quad \mathrm{Negative}\\ \therefore f(x) \, \textrm{is Concave down}\\ \mathrm{Sign\:of\:}f\:''\left(x\right)\mathrm{\:at\:}0<x<\infty \, is \quad \mathrm{Positive}\\ \therefore f(x) \, \textrm{is Concave up}\\ {/eq}

Here, {eq}\;f \textrm{ has inflection points at } x = -\frac{1}{2} \textrm{ and } x = 0. {/eq}


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Concavity and Inflection Points on Graphs

from Math 104: Calculus

Chapter 9 / Lesson 5
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