answer the following questions about the function whose derivative is $f(x)=\\frac{x^{2}(x - 2)}{x + 5}$…

answer the following questions about the function whose derivative is $f(x)=\\frac{x^{2}(x - 2)}{x + 5}$, $x\\neq - 5$.\na. what are the critical points of f?\nb. on what open intervals is f increasing or decreasing?\nc. at what points, if any, does f assume local maximum or minimum values?\na. what are the critical points of f? select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n○ a. the critical point(s) of f is/are $x = \\square$\n(simplify your answer. use a comma to separate answers as needed)\n○ b. the function f has no critical points.

answer the following questions about the function whose derivative is $f(x)=\\frac{x^{2}(x - 2)}{x + 5}$, $x\\neq - 5$.\na. what are the critical points of f?\nb. on what open intervals is f increasing or decreasing?\nc. at what points, if any, does f assume local maximum or minimum values?\na. what are the critical points of f? select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n○ a. the critical point(s) of f is/are $x = \\square$\n(simplify your answer. use a comma to separate answers as needed)\n○ b. the function f has no critical points.

Answer

Explanation:

Step1: Find where (f^{\prime}(x) = 0)

Set (f^{\prime}(x)=\frac{x^{2}(x - 2)}{x + 5}=0). Since a fraction is zero when the numerator is zero (and the denominator is non - zero), we solve (x^{2}(x - 2)=0). Using the zero - product property (a\times b = 0\Rightarrow a = 0) or (b = 0). For (x^{2}(x - 2)=0), we have (x^{2}=0) gives (x = 0) and (x-2=0) gives (x = 2).

Step2: Find where (f^{\prime}(x)) is undefined

The function (y = f^{\prime}(x)) is undefined when the denominator (x + 5=0), so (x=-5). But critical points are in the domain of (f(x)). If (x=-5) is not in the domain of (f(x)) (since (x\neq - 5) for (f^{\prime}(x))), we don't consider it as a critical point.

Answer:

A. The critical point(s) of (f) is/are (x = 0,2)