identify the inflection points and local maxima and minima of the graphed function. identify the open…

identify the inflection points and local maxima and minima of the graphed function. identify the open intervals on which the function is differentiable and is concave up and concave down.\nc. there are no local maxima\nfind each local minimum. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.\na. there is one local minimum value of at\nx=\n(simplify your answers.)\nb. there are two local minima. in increasing order of x - value, the values are and at x = and x =, respectively.\n(simplify your answers.)\nc. there are no local minima.

identify the inflection points and local maxima and minima of the graphed function. identify the open intervals on which the function is differentiable and is concave up and concave down.\nc. there are no local maxima\nfind each local minimum. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.\na. there is one local minimum value of at\nx=\n(simplify your answers.)\nb. there are two local minima. in increasing order of x - value, the values are and at x = and x =, respectively.\n(simplify your answers.)\nc. there are no local minima.

Answer

Explanation:

Step1: Find the first derivative

Given (y = \frac{x^{3}}{3}-x^{2}-3x). Using the power rule ((x^{n})^\prime=nx^{n - 1}), we have (y^\prime=x^{2}-2x - 3). Set (y^\prime = 0), then (x^{2}-2x - 3=(x - 3)(x+1)=0). Solving ((x - 3)(x + 1)=0) gives (x=-1) or (x = 3).

Step2: Use the second - derivative test

Find the second derivative (y^{\prime\prime}=2x-2). When (x=-1), (y^{\prime\prime}(-1)=2\times(-1)-2=-4<0). So (x = - 1) is a local maximum. When (x = 3), (y^{\prime\prime}(3)=2\times3-2 = 4>0). Substitute (x = 3) into the original function (y=\frac{3^{3}}{3}-3^{2}-3\times3=9 - 9-9=-9).

Answer:

A. There is one local minimum value of (-9) at (x = 3).