find the critical points, domain endpoints, and local extreme values for the function. y = x^(2/3)(x - 4) a…

find the critical points, domain endpoints, and local extreme values for the function. y = x^(2/3)(x - 4) a. the critical point(s) or domain endpoint(s) where f is undefined is/are at x = 0. (type an integer or a simplified fraction. use a comma to separate answers as needed.) b. there are no critical points or domain endpoints where f is undefined. what is/are the critical point(s) where f is 0? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the critical point(s) where f is 0 is/are at x = 8/5. (type an integer or a simplified fraction. use a comma to separate answers as needed.) b. there are no critical points where f is 0. from the critical point(s) and domain endpoint(s), what is/are the point(s) corresponding to local maxima? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the point(s) corresponding to the local maxima is/are (0,0). (type an ordered pair. use integers or decimals for any numbers in the expression. round to the nearest thousandth as needed. use a comma to separate answers as needed.) b. there are no points corresponding to local maxima. from the critical point(s) and domain endpoint(s), what is/are the point(s) corresponding to local minima? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the point(s) corresponding to the local minima is/are. (type an ordered pair. use integers or decimals for any numbers in the expression. round to the nearest thousandth as needed. use a comma to separate answers as needed.) b. there are no points corresponding to local minima.

find the critical points, domain endpoints, and local extreme values for the function. y = x^(2/3)(x - 4) a. the critical point(s) or domain endpoint(s) where f is undefined is/are at x = 0. (type an integer or a simplified fraction. use a comma to separate answers as needed.) b. there are no critical points or domain endpoints where f is undefined. what is/are the critical point(s) where f is 0? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the critical point(s) where f is 0 is/are at x = 8/5. (type an integer or a simplified fraction. use a comma to separate answers as needed.) b. there are no critical points where f is 0. from the critical point(s) and domain endpoint(s), what is/are the point(s) corresponding to local maxima? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the point(s) corresponding to the local maxima is/are (0,0). (type an ordered pair. use integers or decimals for any numbers in the expression. round to the nearest thousandth as needed. use a comma to separate answers as needed.) b. there are no points corresponding to local maxima. from the critical point(s) and domain endpoint(s), what is/are the point(s) corresponding to local minima? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the point(s) corresponding to the local minima is/are. (type an ordered pair. use integers or decimals for any numbers in the expression. round to the nearest thousandth as needed. use a comma to separate answers as needed.) b. there are no points corresponding to local minima.

Answer

Explanation:

Step1: Find the derivative

First, use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = x^{2/3}$ and $v=x - 4$. The derivative of $u=x^{2/3}$ is $u^\prime=\frac{2}{3}x^{-1/3}$, and the derivative of $v=x - 4$ is $v^\prime = 1$. So $y^\prime=\frac{2}{3}x^{-1/3}(x - 4)+x^{2/3}\times1=\frac{2(x - 4)}{3x^{1/3}}+x^{2/3}=\frac{2x-8 + 3x}{3x^{1/3}}=\frac{5x-8}{3x^{1/3}}$.

Step2: Find where $y^\prime$ is undefined

The derivative $y^\prime=\frac{5x - 8}{3x^{1/3}}$ is undefined when the denominator $3x^{1/3}=0$, which gives $x = 0$.

Step3: Find where $y^\prime$ is 0

Set the numerator of $y^\prime$ equal to 0: $5x-8 = 0$. Solving for $x$ gives $x=\frac{8}{5}$.

Step4: Use the first - derivative test to find local extrema

We consider the intervals $(-\infty,0)$, $(0,\frac{8}{5})$, and $(\frac{8}{5},\infty)$.

  • For $x\in(-\infty,0)$, choose $x=-1$. Then $y^\prime=\frac{5\times(-1)-8}{3\times(-1)^{1/3}}=\frac{-13}{-3}=\frac{13}{3}>0$.
  • For $x\in(0,\frac{8}{5})$, choose $x = 1$. Then $y^\prime=\frac{5\times1-8}{3\times1^{1/3}}=\frac{-3}{3}=-1<0$.
  • For $x\in(\frac{8}{5},\infty)$, choose $x = 2$. Then $y^\prime=\frac{5\times2-8}{3\times2^{1/3}}=\frac{2}{3\times2^{1/3}}>0$.

Since the function changes from increasing to decreasing at $x = 0$, $y(0)=0^{2/3}(0 - 4)=0$ is a local maximum. Since the function changes from decreasing to increasing at $x=\frac{8}{5}$, $y(\frac{8}{5})=(\frac{8}{5})^{2/3}(\frac{8}{5}-4)=(\frac{8}{5})^{2/3}\times(-\frac{12}{5})\approx - 3.175$ is a local minimum.

Answer:

The point(s) corresponding to the local minima is/are $(\frac{8}{5}, - 3.175)$