Question

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

Answer 1

a) The critical points are
`x = 3` and
`x = -6`.

b) f is decreasing in the interval
`(-\infty, -6)`

f is increasing in the intervals
`(-6,3)` and
`(3,\infty)`.

c) Local minima:
`x = -6`

Local maxima: No local maxima

Explanation:

(a) what are the critical points of f?

The critical points of f are those in which
`f^(\prime)(x) = 0`. So

`f^(\prime)(x) = 0`

`(x-3)^(2)(x+6) = 0`

So, the critical points are
`x = 3` and
`x = -6`.

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

For any interval, if
`f^(\prime)` is positive, f is increasing in the interval. If it is negative, f is decreasing in the interval.

Our critical points are
`x = 3` and
`x = -6`. So we have those following intervals:

`(-\infty, -6), (-6,3), (3, \infty)`

We select a point x in each interval, and calculate
`f^(\prime)(x)`.

So

-------------------------

`(-\infty, -6)`

`f^(\prime)(-7) = (-7-3)^(2)(-7+6) = (100)(-1) = -100`

f is decreasing in the interval
`(-\infty, -6)`

---------------------------

`(-6,3)`

`f^(\prime)(2) = (2-3)^(2)(2+6) = (1)(8) = 8`

f is increasing in the interval
`(-6,3)`.

------------------------------

`(3, \infty)`

`f^(\prime)(4) = (4-3)^(2)(4+6) = (1)(10) = 10`

f is increasing in the interval
`(3,\infty)`.

(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
`x = -6`, f goes from decreasing to increasing.

So
`x = -6`, f assume a local minima value

At
`x = 3`, 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.