# Difference between revisions of "009A Sample Final 3, Problem 6"

Let

${\displaystyle f(x)=4+8x^{3}-x^{4}}$

(a) Over what  ${\displaystyle x}$-intervals is  ${\displaystyle f}$  increasing/decreasing?

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

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

(d) Sketch the shape of the graph of  ${\displaystyle f.}$

Foundations:
1. ${\displaystyle f(x)}$  is increasing when  ${\displaystyle f'(x)>0}$  and  ${\displaystyle f(x)}$  is decreasing when  ${\displaystyle f'(x)<0.}$
2. The First Derivative Test tells us when we have a local maximum or local minimum.
3. ${\displaystyle f(x)}$  is concave up when  ${\displaystyle f''(x)>0}$  and  ${\displaystyle f(x)}$  is concave down when  ${\displaystyle f''(x)<0.}$

Solution:

(a)

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

(b)

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

(c)

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

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