009A Sample Final 1, Problem 9

Given the function $f(x)=x^{3}-6x^{2}+5$,

a) Find the intervals in which the function increases or decreases.

b) Find the local maximum and local minimum values.

c) Find the intervals in which the function concaves upward or concaves downward.

d) Find the inflection point(s).

e) Use the above information (a) to (d) to sketch the graph of $y=f(x)$.

Foundations:

Solution:

(a)

Step 1:
We start by taking the derivative of $f(x)$. We have $f'(x)=3x^{2}-12x$.
Now, we set $f'(x)=0$. So, we have $0=3x(x-4)$.
Hence, we have $x=0$ and $x=4$.
So, these values of $x$ break up the number line into 3 intervals: $(-\infty ,0),(0,4),(4,\infty )$.

Step 2:
To check whether the function is increasing or decreasing in these intervals, we use testpoints.
For $x=-1,~f'(x)=15>0$.
For $x=1,~f'(x)=-9<0$.
For $x=5,~f'(x)=15>0$.
Thus, $f(x)$ is increasing on $(-\infty ,0),(4,\infty )$ and decreasing on Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (0,4)} .

(b)

Step 1:
The local maximum occurs at $x=0$ and the local minimum occurs at $x=4$.

Step 2:
So, the local maximum value is $f(0)=5$ and the local minimum value is $f(4)=-27$.

(c)

Step 1:
To find the intervals when the function is concave up or concave down, we need to find $f''(x)$.
We have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f''(x)=6x-12} .
We set $f''(x)=0$.
So, we have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 0=6x-12} . Hence, $x=2$.
This value breaks up the number line into two intervals: Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (-\infty ,2),(2,\infty )} .

Step 2:
Again, we use test points in these two intervals.
For $x=0$, we have $f''(x)=-12<0$.
For $x=3$, we have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f''(x)=6>0} .
Thus, $f(x)$ is concave up on the interval $(2,\infty )$ and concave down on the interval $(-\infty ,2)$.

(d)

Step 1:
Using the information from part (c), there is one inflection point that occurs at $x=2$.
Now, we have $f(2)=8-24+5=-11$.
So, the inflection point is $(2,-11)$.

(e)

Step 1:
Insert sketch here.

Final Answer:
(a) $f(x)$ is increasing on $(-\infty ,0),(4,\infty )$ and decreasing on Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (0,4)} .
(b) The local maximum value is $f(0)=5$ and the local minimum value is $f(4)=-27$.
(c) $f(x)$ is concave up on the interval $(2,\infty )$ and concave down on the interval $(-\infty ,2)$.
(d) $(2,-11)$
(e) See Step 1 for graph.

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