Answer the following question about the function whose derivative is given below

f'(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

a) The critical points are and .

b) f is decreasing in the interval

f is increasing in the intervals and .

c) Local minima:

Local maxima: No local maxima

Step-by-step explanation:

(a) what are the critical points of f?

The critical points of f are those in which . So

So, the critical points are and .

(b) on what intervals is f increasing or decreasing? (if there is no interval put no interval)

For any interval, if is positive, f is increasing in the interval. If it is negative, f is decreasing in the interval.

Our critical points are and . So we have those following intervals:

We select a point x in each interval, and calculate .

So

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f is decreasing in the interval

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f is increasing in the interval .

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f is increasing in the 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

At 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 , f goes from decreasing to increasing.

So , f assume a local minima value

At , 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.

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23