Question

consider the following function. (if an answer does not exist, enter dne.)
$f(x)=x^{3}-12x + 4$
(a) find the interval of increase. (enter your answer using interval notation.)
find the interval of decrease. (enter your answer using interval notation.)
(b) find the local minimum value(s). (enter your answers as a comma - separated list.)
find the local maximum value(s). (enter your answers as a comma - separated list.)
(c) find the inflection point.
$(x,y)=(\square)$
find the interval where the graph is concave upward. (enter your answer using interval notation.)
find the interval where the graph is concave downward. (enter your answer using interval notation.)

Explanation:

Step1: Find the first derivative

The derivative of \(f(x)=x^{3}-12x + 4\) is \(f^\prime(x)=3x^{2}-12\).
Set \(f^\prime(x)=0\), then \(3x^{2}-12 = 0\), \(x^{2}=4\), \(x=\pm2\).

Step2: Determine intervals of increase and decrease

For \(x\in(-\infty,-2)\), let \(x=-3\), \(f^\prime(-3)=3\times(-3)^{2}-12=15>0\).
For \(x\in(-2,2)\), let \(x = 0\), \(f^\prime(0)=3\times0^{2}-12=-12<0\).
For \(x\in(2,\infty)\), let \(x = 3\), \(f^\prime(3)=3\times3^{2}-12=15>0\).
So the interval of increase is \((-\infty,-2)\cup(2,\infty)\), and the interval of decrease is \((-2,2)\).

Step3: Find local minima and maxima

Since \(f(x)\) changes from increasing to decreasing at \(x=-2\), \(f(-2)=(-2)^{3}-12\times(-2)+4=20\) (local maximum).
Since \(f(x)\) changes from decreasing to increasing at \(x = 2\), \(f(2)=2^{3}-12\times2+4=-12\) (local minimum).

Step4: Find the second derivative

\(f^{\prime\prime}(x)=6x\).
Set \(f^{\prime\prime}(x)=0\), then \(x = 0\).
When \(x = 0\), \(y=f(0)=4\). So the inflection point is \((0,4)\).
For \(x\in(0,\infty)\), \(f^{\prime\prime}(x)>0\) (concave upward).
For \(x\in(-\infty,0)\), \(f^{\prime\prime}(x)<0\) (concave downward).

Answer:

(a) Interval of increase: \((-\infty,-2)\cup(2,\infty)\); Interval of decrease: \((-2,2)\)
(b) Local minimum value: \(-12\); Local maximum value: \(20\)
(c) Inflection point: \((0,4)\); Concave upward interval: \((0,\infty)\); Concave downward interval: \((-\infty,0)\)