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Find all the inflection numbers, and determine the interval (interval notation) where the...

Question:

Find all the inflection numbers, and determine the interval (interval notation) where the function is concave upward or concave downward.

{eq}\displaystyle f(x) = x^{\dfrac 8 3} + x^{\dfrac 5 3} {/eq}.

Concavity and Inflection Points:

The second derivative of a function {eq}f(x) {/eq} allows us to determine information about the graph of {eq}f(x) {/eq}.

For the intervals on which {eq}f''(x)>0 {/eq}, the graph of {eq}f(x) {/eq} is concave up.

For the intervals on which {eq}f''(x)<0 {/eq}, the graph of {eq}f(x) {/eq} is concave down.

If {eq}f''(x) {/eq} changes signs when {eq}x=a {/eq}, {eq}f(x) {/eq} has a point of inflection when {eq}x=a {/eq}.

Answer and Explanation:

In order to find the inflection numbers and intervals of concavity for {eq}\displaystyle f(x) = x^{\dfrac 8 3} + x^{\dfrac 5 3} {/eq}, we need to find its second derivative.


{eq}{f}'\left ( x \right )=\frac{8}{3}x^{\frac{5}{3}}+\frac{5}{3}x^{\frac{2}{3}} {/eq}

{eq}\begin{align*} {f}''\left ( x \right )&=\frac{40}{9}x^{\frac{2}{3}}+\frac{10}{9}x^{-\frac{1}{3}}\\ &=\frac{40x^{\frac{2}{3}}}{9}+\frac{10}{9x^{\frac{1}{3}}}\\ &=\frac{40x^{\frac{2}{3}}}{9}\left ( \frac{x^{\frac{1}{3}}}{x^{\frac{1}{3}}} \right )+\frac{10}{9x^{\frac{1}{3}}}\\ &=\frac{40x+10}{9x^{\frac{1}{3}}} \end{align*} {/eq}


{eq}f''(x)=0 {/eq} when {eq}x=-\frac{1}{4} {/eq}. {eq}f''(x) {/eq} does not exist when {eq}x=0 {/eq}.

We will examine {eq}f''(x) {/eq} on intervals {eq}(-\infty,-\frac{1}{4}),(-\frac{1}{4},0) {/eq}, and {eq}(0,\infty) {/eq}.


{eq}{f}''\left ( -1 \right )=\displaystyle{\frac{40\left ( -1 \right )+10}{9\left ( -1 \right )^{\frac{1}{3}}}>0} {/eq}

So {eq}f''(x)>0 {/eq} on {eq}(-\infty,-\frac{1}{4}) {/eq}.

{eq}{f}''\left ( \frac{-1}{8} \right )=\displaystyle{\frac{40\left ( \frac{-1}{8} \right )+10}{9\left ( \frac{-1}{8} \right )^{\frac{1}{3}}}<0} {/eq}

So {eq}f''(x)<0 {/eq} on {eq}(-\frac{1}{4},0) {/eq}.

{eq}{f}''\left ( 1 \right )=\displaystyle{\frac{40\left ( 1 \right )+10}{9\left ( 1\right )^{\frac{1}{3}}}>0} {/eq}

So {eq}f''(x)>0 {/eq} on {eq}(0,\infty) {/eq}.


Since {eq}f''(x)>0 {/eq} on {eq}(-\infty,-\frac{1}{4})\cup \left ( 0,\infty \right ) {/eq}, the graph of {eq}f(x) {/eq} is concave up on {eq}\boxed{(-\infty,-\frac{1}{4})\cup \left ( 0,\infty \right )} {/eq}.

Since {eq}f''(x)<0 {/eq} on {eq}(-\frac{1}{4},0) {/eq}, the graph of {eq}f(x) {/eq} is concave down on {eq}\boxed{(-\frac{1}{4},0)} {/eq}.


Since {eq}f''(x) {/eq} changes signs when {eq}x=-\frac{1}{4} {/eq} and when {eq}x=0 {/eq}, the graph of {eq}f(x) {/eq} has inflection points when {eq}\boxed{x=-\frac{1}{4}} {/eq} and when {eq}\boxed{x=0} {/eq}.


Learn more about this topic:

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Understanding Concavity and Inflection Points with Differentiation

from Math 104: Calculus

Chapter 10 / Lesson 6
14K

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