How do you find all intervals where the function $f(x)=1/3x^3+3/2x^2+2$ is increasing?

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Answer 1 · 2014-09-25T20:29:46

$f$ is increasing on $(-infty,-3]$ and $[0,infty)$ .

The graph of $f$ looks like this:

enter image source here

Let us look at some details.

$f(x)=1/3x^3+3/2x^2+2$

By solving $f'(x)=0$ for $x$ ,

$f'(x)=x^2+3x=x(x+3)=0$ ,

we find the critical values: $x=-3, 0$ .

Using the critical values to divide $(-infty,infty)$ into

$(-infty,-3]$ , $[-3,0]$ , and $[0,infty)$

and we choose sample values $-4$ , $-2$ and $1$ for the intervals above, respectively. (We can use any number on each interval excluding endpoints.)

$f'(-4)=4>0 Rightarrow f$ is increasing on $(-infty,-3]$

$f'(-2)=-2<0 Rightarrow f$ is decreasing on $[-3,0]$

$f'(1)=4>0 Rightarrow f$ is increasing on $[0,infty)$

How do you find all intervals where the function f(x)=1/3x^3+3/2x^2+2f(x)=1/3x^3+3/2x^2+2 is increasing?