question 6 of 16\nview policies\nshow attempt history\ncurrent attempt in progress\nyour answer is partially…

question 6 of 16\nview policies\nshow attempt history\ncurrent attempt in progress\nyour answer is partially correct.\nfind all critical points and then use the first - derivative test to determine local maxima and minima. check your answer by gr\n f(x)=(x^{2}-16)^{7} \nenter the critical points in increasing order.\nif there is no local maximum or local minimum, enter na.\nx = - 4\nx = 0\nx = 4\nthe local maximum is at x =\nthe local minimum is at x =\netextbook and media\nhint

question 6 of 16\nview policies\nshow attempt history\ncurrent attempt in progress\nyour answer is partially correct.\nfind all critical points and then use the first - derivative test to determine local maxima and minima. check your answer by gr\n f(x)=(x^{2}-16)^{7} \nenter the critical points in increasing order.\nif there is no local maximum or local minimum, enter na.\nx = - 4\nx = 0\nx = 4\nthe local maximum is at x =\nthe local minimum is at x =\netextbook and media\nhint

Answer

Explanation:

Step1: Find the derivative

Use the chain - rule. If $y = u^{7}$ and $u=x^{2}-16$, then $\frac{dy}{du}=7u^{6}$ and $\frac{du}{dx}=2x$. So, $f^\prime(x)=7(x^{2}-16)^{6}\cdot2x = 14x(x^{2}-16)^{6}=14x(x - 4)^{6}(x + 4)^{6}$.

Step2: Find critical points

Set $f^\prime(x)=0$. Since $(x - 4)^{6}\geq0$ and $(x + 4)^{6}\geq0$ for all real $x$, then $14x(x - 4)^{6}(x + 4)^{6}=0$ when $x=-4,0,4$.

Step3: Use the first - derivative test

Choose test points in the intervals $(-\infty,-4)$, $(-4,0)$, $(0,4)$ and $(4,\infty)$.

  • For $x=-5$ (in the interval $(-\infty,-4)$), $f^\prime(-5)=14\times(-5)\times((-5)^{2}-16)^{6}<0$.
  • For $x = - 1$ (in the interval $(-4,0)$), $f^\prime(-1)=14\times(-1)\times((-1)^{2}-16)^{6}<0$.
  • For $x = 1$ (in the interval $(0,4)$), $f^\prime(1)=14\times1\times(1^{2}-16)^{6}>0$.
  • For $x = 5$ (in the interval $(4,\infty)$), $f^\prime(5)=14\times5\times(5^{2}-16)^{6}>0$.

Since the function is decreasing on $(-\infty,0)$ and increasing on $(0,\infty)$, the local minimum is at $x = 0$. There is no local maximum.

Answer:

The local maximum is at $x =$ NA The local minimum is at $x = 0$