Homework Help: Minimum and Maximum Values

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1. Feb 24, 2017

a1234

1. The problem statement, all variables and given/known data

The problem is in the attached file. The part I need a little help with is part b.

2. Relevant equations and attempt at a solution

For part a, I got h(8) = 2, h'(6) = -2, and h''(4) = -2.
For part c, I found that the integral from 0 to 5 is 7, so I multiplied 7 by 7 to get 49 (because 35 = 5*7). And f(5) = 0, so f(108) also equals 0. I got the tangent line y - 49 = 0(x - 35).

For part b, I think we're supposed to consider x = 0 and x = 3, since this is where the graph starts increasing from decreasing and vice versa. But I also think we have to evaluate the integral at these points. Can someone explain this?

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2. Feb 24, 2017

haruspex

The period is not 5, so why is that interesting?

3. Feb 24, 2017

BvU

If I consider
as an attempt, I agree with 0 and 3 but I don't see it decrease before 0... I also see another interval where f is not bigger than at 0 ...

If you want help for part (a) -- and I think you need it -- , please post your working in detail.

 strike because the question is about $h$ and I mistakenly read $f$ --- bedtime

4. Feb 24, 2017

haruspex

Did you make a similar mistake here? I agree with all the answers for a).

5. Feb 24, 2017

BvU

YES !! aaaarrghh

6. Feb 24, 2017

a1234

I thought the graph would keep repeating every 5 units, but that's not true.
It seems to repeat every 8 units, so could we do this?

integral from 0 to 8 = 1
8*4 = 32
1*4 = 4
integral from 0 to 3 = 3
4 + 3 = 7
y - 7 = 0(x - 35)?

7. Feb 24, 2017

haruspex

You are told it does.
Didn"t you calculate it as 2 in part a?
I do not understand that step. What is the gradient of h at x=35?

8. Feb 25, 2017

a1234

I didn't count the triangle from 0 to 1 when I first calculated the integral from 0 to 8. I think it should be 1 because the area of the trapezoid is -6, area of the large triangle is 8, and area of the small triangle is -1. So 8 + (-6) + (-1) = 1.

For the derivative at x = 35...
h(35) = integral from 0 to 35 = 7
h'(35) = f(35) = 4 because the graph repeats every 8 units...f(3) = 4 for the remaining 3 units after 8*4 = 32. So the derivative at x = 35 is 4.
If I put it into point-slope form, y - 7 = 4(x - 35) is the tangent line.

9. Feb 25, 2017

haruspex

The trapezoid extends to x=9.

10. Feb 25, 2017

a1234

Ugh. I'm making a lot of mistakes here.

You're right, we only need the area of the trapezoid up to x = 8. So then the area of the trapezoid is 5, and h(8) = 2. The integral from 0 to 35 is 2*4 + 3 = 11.
y - 11 = 4(x - 35)

11. Feb 25, 2017

haruspex

Looks right.

12. Feb 25, 2017

a1234

How do I go about part b?

13. Feb 25, 2017

haruspex

Do you understand about local extrema and absolute extrema?
How do you find a local extremum?

14. Feb 26, 2017

a1234

The absolute maximum is the highest point over the entire graph. The absolute minimum is the lowest point over the graph. The relative max/min is the maximum or minimum over a certain interval.

I think the first step is to find do h'(x), which is just f(x) and find the critical points and endpoints of the graph. The critical points are 1 and 5, and the endpoints are 0 and 7. And then we need to evaluate g at those values. I came up with the following:
h(0) = 0
h(1) = -1
h(5) = 7
h(7) = 4
The absolute min is -1 and max is 7. The relative min is 0 and max is 4.

15. Feb 26, 2017

haruspex

You are asked for the absolute min and max over the whole interval, so yes, -1 and 7.

16. Feb 26, 2017

a1234

Okay. Thank you very much for the help!