Mathematics, 16.10.2019 01:20 morganluvsblueow9wtm

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

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Answer from: Jasonn7787

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

Step-by-step 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 $f^{\prime}(x)$ .

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

$(-\infty, -6)$

$f^{\prime}(-7) = (-7-3)^{2}(-7+6) = (100)(-1) = -100$

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.

Answer from: Quest

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

Answer from: Quest

M< 5 = 80 , < 2 = 80, < 6= 100, < 3= 100, < 7= 80, < 4= 80, < 8= 100

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