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

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

Answer from: Quest

answer: 8d

step-by-step explanation:

45+5= 50+5=55+5=60