Answer the following question about the function whose derivative is given belowf'(x)=(x-3)^2(x+6)(a) what are the critical points of f?(b) on what intervals is f increasing or decreasing? (if there is no interval put no interval)(c) At what points, if any, does f assume local maximum and minima values. ( if there is no local maxima put mo local maxima) if there is no local minima put no local minima

Answer:a) The critical points are [tex]x = 3[/tex] and [tex]x = -6[/tex].b) f is decreasing in the interval [tex](-\infty, -6)[/tex]f is increasing in the intervals [tex](-6,3)[/tex] and [tex](3,\infty)[/tex].c) Local minima: [tex]x = -6[/tex]Local maxima: No local maximaStep-by-step explanation:(a) what are the critical points of f?The critical points of f are those in which [tex]f^{\prime}(x) = 0[/tex]. So[tex]f^{\prime}(x) = 0[/tex][tex](x-3)^{2}(x+6) = 0[/tex]So, the critical points are [tex]x = 3[/tex] and [tex]x = -6[/tex].(b) on what intervals is f increasing or decreasing? (if there is no interval put no interval)For any interval, if [tex]f^{\prime}[/tex] is positive, f is increasing in the interval. If it is negative, f is decreasing in the interval. Our critical points are [tex]x = 3[/tex] and [tex]x = -6[/tex]. So we have those following intervals:[tex](-\infty, -6), (-6,3), (3, \infty)[/tex]We select a point x in each interval, and calculate [tex]f^{\prime}(x)[/tex].So-------------------------[tex](-\infty, -6)[/tex][tex]f^{\prime}(-7) = (-7-3)^{2}(-7+6) = (100)(-1) = -100[/tex]f is decreasing in the interval [tex](-\infty, -6)[/tex]---------------------------[tex](-6,3)[/tex][tex]f^{\prime}(2) = (2-3)^{2}(2+6) = (1)(8) = 8[/tex]f is increasing in the interval [tex](-6,3)[/tex].------------------------------[tex](3, \infty)[/tex][tex]f^{\prime}(4) = (4-3)^{2}(4+6) = (1)(10) = 10[/tex]f is increasing in the interval [tex](3,\infty)[/tex].(c) At what points, if any, does f assume local maximum and minima values. ( if there is no local maxima put mo local maxima) if there is no local minima put no local minimaAt a critical point x, if the function goes from decreasing to increasing, it is a local minima. And if the function goes from increasing to decreasing, it is a local maxima.So, for each critical point is this problem:At [tex]x = -6[/tex], f goes from decreasing to increasing.So [tex]x = -6[/tex], f assume a local minima valueAt [tex]x = 3[/tex], f goes from increasing to increasing. So, there it is not a local maxima nor a local minima. So, there is no local maxima for this function.