009A Sample Final 3, Problem 6

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:

Step 2:

(d):
Insert graph

Final Answer:
(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)
(d)    See above

Return to Sample Exam