consider the following function. (if an answer does not exist, enter dne.)\n$f(x)=x^{3}-12x + 4$\n(a) find…

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))