# 009A Sample Final 3, Problem 6

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Let

$f(x)=4+8x^{3}-x^{4}$ (a) Over what  $x$ -intervals is  $f$ increasing/decreasing?

(b) Find all critical points of  $f$ and test each for local maximum and local minimum.

(c) Over what  $x$ -intervals is  $f$ concave up/down?

(d) Sketch the shape of the graph of  $f.$ Foundations:
1. $f(x)$ is increasing when  $f'(x)>0$ and  $f(x)$ is decreasing when  $f'(x)<0.$ 2. The First Derivative Test tells us when we have a local maximum or local minimum.
3. $f(x)$ is concave up when  $f''(x)>0$ and  $f(x)$ is concave down when  $f''(x)<0.$ Solution:

(a)

Step 1:
We start by taking the derivative of  $f(x).$ We have  $f'(x)=24x^{2}-4x^{3}.$ Now, we set  $f'(x)=0.$ So, we have
$0=4x^{2}(6-x).$ Hence, we have  $x=0$ and  $x=6.$ So, these values of  $x$ break up the number line into 3 intervals:
$(-\infty ,0),(0,6),(6,\infty ).$ Step 2:
To check whether the function is increasing or decreasing in these intervals, we use testpoints.
For  $x=-1,~f'(x)=28>0.$ For  $x=1,~f'(x)=20>0.$ For  $x=7,~f'(x)=-196<0.$ Thus,  $f(x)$ is increasing on  $(-\infty ,6)$ and decreasing on  $(6,\infty ).$ (b)

Step 1:
The critical points of  $f(x)$ occur at  $x=0$ and  $x=6.$ Plugging these values into  $f(x),$ we get the critical points
$(0,4)$ and  $(6,436).$ Step 2:
Using the first derivative test and the information from part (a),
$(0,4)$ is not a local minimum or local maximum and
$(6,436)$ is a local maximum.

(c)

Step 1:
To find the intervals when the function is concave up or concave down, we need to find  $f''(x).$ We have  $f''(x)=48x-12x^{2}.$ We set  $f''(x)=0.$ So, we have
$0=12x(4-x).$ Hence,  $x=0$ and  $x=4$ .
This value breaks up the number line into three intervals:
$(-\infty ,0),(0,4),(4,\infty ).$ Step 2:
Again, we use test points in these three intervals.
For  $x=-1,$ we have  $f''(x)=-60<0.$ For  $x=1,$ we have  $f''(x)=48>0.$ For  $x=5,$ we have  $f''(x)=-60<0.$ Thus,  $f(x)$ is concave up on the interval  $(0,4)$ and concave down on the interval  $(-\infty ,0)\cup (4,\infty ).$ (d):
Insert graph

(a)    $f(x)$ is increasing on  $(-\infty ,6)$ and decreasing on  $(6,\infty ).$ (b)    The critical points are $(0,4)$ and  $(6,436).$ $(0,4)$ is not a local minimum or local maximum and  $(6,436)$ is a local maximum.
(c)   $f(x)$ is concave up on the interval  $(0,4)$ and concave down on the interval  $(-\infty ,0)\cup (4,\infty ).$ 