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Find the intervals on which the graph of f is concave upward, and the intervals on which the...

Question:

Find the intervals on which the graph of {eq}f {/eq} is concave upward, and the intervals on which the graph of {eq}f {/eq} is concave downward, and the x coordinates of the inflection points.

{eq}f(x)=-x^3-5x^2+4x-3 {/eq}

Concavity and Inflection Points:

Given a function {eq}f(x) {/eq}, where the second derivative of the function is positive the graph of the function is concave upward. Where the second derivative of the function is negative the graph of the function is concave downward. Inflection points occur where the graph of the function changes from concave upward to concave downward.

Answer and Explanation:

Step 1. Find {eq}f'(x) {/eq}.

{eq}\begin{align} \displaystyle f(x)\displaystyle &=-x^3-5x^2+4x-3\\ f'(x)\displaystyle &=-3x^{2}-10x+4\\ \end{align} {/eq}

Step 2. Find {eq}f\:''(x) {/eq}.

{eq}\begin{align} \displaystyle f'(x)\displaystyle &=-3x^{2}-10x+4\\ f\:''(x)\displaystyle &=-6x-10\\ \end{align} {/eq}

Step 3. Find the intervals where {eq}f\:''(x)>0 {/eq} and {eq}f\:''(x)<0 {/eq}.

Factor {eq}f\:''(x) {/eq}.

{eq}\begin{align} \displaystyle f\:''(x)\displaystyle &=-6x-10\\ \displaystyle &=-2(3x+5)\\ \end{align} {/eq}

{eq}\begin{align} \displaystyle -2<0\hspace{10mm}3x+5<0\:for\:x:\:x\displaystyle &<-\frac{5}{3}\\ 3x+5>0\:for\:x:\:x\displaystyle &>-\frac{5}{3}\\ \end{align} {/eq}

{eq}\left \{ x:\:x<-\frac{5}{3} \right \}\hspace{3mm}f\:''(x)=(negative)(negative)>0\\ \left \{ x:\:x>-\frac{5}{3} \right \}\hspace{3mm}f\:''(x)=(negative)(positive)<0\\ {/eq}

Therefore {eq}f(x) {/eq} is concave upward on the interval {eq}\left(-\infty,\:-\frac{5}{3}\right) {/eq} and is concave downward on the interval {eq}\left(-\frac{5}{3},\:\infty\right) {/eq}.

There is an inflection point at {eq}x=-\frac{5}{3} {/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
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