# 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.

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}.