let f(x)=8x^4(x - 6)^2. (a) use interval notation to indicate where f(x) is increasing. note: use inf for ∞…

Explanation:

Step1: Find the derivative of $f(x)$

Use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = 8x^{4}$ and $v=(x - 6)^{2}=x^{2}-12x + 36$. $u^\prime=32x^{3}$ and $v^\prime = 2(x - 6)=2x-12$. $f^\prime(x)=32x^{3}(x - 6)^{2}+8x^{4}\times2(x - 6)$ $=8x^{3}(x - 6)[4(x - 6)+2x]$ $=8x^{3}(x - 6)(4x-24 + 2x)$ $=8x^{3}(x - 6)(6x-24)$ $=48x^{3}(x - 6)(x - 4)$.

Step2: Find the critical points

Set $f^\prime(x)=0$. $48x^{3}(x - 6)(x - 4)=0$. The critical points are $x = 0,x = 4,x = 6$.

Step3: Test the intervals

Choose test - points in the intervals $(-\infty,0),(0,4),(4,6),(6,\infty)$. For $x=-1$: $f^\prime(-1)=48\times(-1)^{3}\times(-1 - 6)\times(-1 - 4)=48\times(-1)\times(-7)\times(-5)<0$. For $x = 1$: $f^\prime(1)=48\times1^{3}\times(1 - 6)\times(1 - 4)=48\times1\times(-5)\times(-3)>0$. For $x = 5$: $f^\prime(5)=48\times5^{3}\times(5 - 6)\times(5 - 4)=48\times125\times(-1)\times1<0$. For $x = 7$: $f^\prime(7)=48\times7^{3}\times(7 - 6)\times(7 - 4)=48\times343\times1\times3>0$.

Step4: Determine increasing and decreasing intervals

(A) Increasing intervals: $(0,4)\cup(6,\infty)$. (B) Decreasing intervals: $(-\infty,0)\cup(4,6)$.

Step5: Identify local maxima and minima

Since $f(x)$ changes from decreasing to increasing at $x = 0$ and $x = 6$, these are local minima. Since $f(x)$ changes from increasing to decreasing at $x = 4$, this is a local maximum.

Step6: Calculate the averages

(C) The $x$ - value of the local maximum is $x = 4$, so the average of the $x$ - values of all local maxima is $4$. (D) The $x$ - values of the local minima are $x = 0$ and $x = 6$, and the average is $\frac{0 + 6}{2}=3$.

Answer:

(A) $(0,4)\cup(6,\infty)$ (B) $(-\infty,0)\cup(4,6)$ (C) $4$ (D) $3$