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Revision as of 23:21, 4 March 2016
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:

Recall:

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 $f''(x)<0.$

4. Inflection points occur when $f''(x)=0.$

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 $(\infty ,0),(0,4),(4,\infty ).$$0=3x(x4).$

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 $(\infty ,0)\cup (4,\infty ),$ and decreasing on $(0,4).$

(b)
Step 1:

By the First Derivative Test, 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 $f''(x)=6x12.$

We set $f''(x)=0.$

So, we have $0=6x12.$ Hence, $x=2.$

This value breaks up the number line into two intervals: $(\infty ,2),(2,\infty ).$

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 $f''(x)=6>0.$

Thus, $f(x)$ is concave up on the interval $(2,\infty ),$ and concave down on the interval $(\infty ,2).$

(d)

Using the information from part (c), there is one inflection point that occurs at $x=2.$

Now, we have $f(2)=824+5=11.$

So, the inflection point is $(2,11).$

Final Answer:

(a) $f(x)$ is increasing on $(\infty ,0),(4,\infty ),$ and decreasing on $(0,4).$

(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 graph in (e).

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