Question

Find Relative Extrema for:

f(x)=1/4 (x+2)(x-1)^2

h(x)=2x^3+5x^2-25x

Answer 1

Calculate f'(x) and see when f'(x)=0. If f''(x1)<0 then x1 is a maxima. If f''(x2)>0 then x2 is a minima.

Answer 2

I'm not even sure what the relative extreme answer should look like

Answer 3

The answer will be an ordered pair such as: (a, b). When x = a, either a minimum or a maximum of f(x) occurs and b is the corresponding minimum or maximum value.

Answer 4

Ok, but I still don't know how to solve this. Could you help?

Answer 5

f(x)=1/4 (x+2)(x-1)^2
Find the derivative of f(x).
f'(x) = ?

Answer 6

How do I do that

Answer 7

You have not been taught differentiation or finding the derivative yet?

Answer 8

I probably have, but I must not remember

Answer 9

This may not be the suitable place to teach how to do differentiation. If you have a book you can go through the basics first such as the power rule, the chain rule, the product rule and the quotient rule. Then if you have doubts you can ask and one of us will be able to assist you. But teaching calculus here for the first time via typing will be too time consuming.

Answer 10

Is the derivative 3x^2-3 / 4

Answer 11

It is 3/4(x - 1)(x + 1) or 3/4x^2 - 3/4.

Answer 12

Ok so now we know the derivative, how do I find the relative extrema?

Answer 13

Equate it to 0 and solve for x.
If you leave the derivative in the factored form: 3/4(x - 1)(x + 1) it will be easier to find the zeros.

Answer 14

x=1,-1

Answer 15

Yes, those are the x values where a relative minimum or maximum occurs.

Put x = 1 and find f(1)
Put x = -1 and find f(-1)

Answer 16

(0,1) is the relative extrema?

Answer 17

The first value should be the x. The second should be the y.

Answer 18

hmm I'm not sure. When I plugged 1 in I got 0 and when I put -1 I got 1

Answer 19

when x = 1, y = 0. So the point is (1, 0)
when x = -1, y = 1. So the other point is (-1, 1)

Before concluding these are the extrema (meaning maxima and minima) you should find the second derivative and find evaluate f''(x) at x = 1 and -1.

f'(x) = 3/4x^2 - 3/4
f''(x) = ?

Answer 20

when fx=1, x=1
when fx=-1, x=-1

Answer 21

Not sure what that is.

Answer 22

When x = 1, f(x) = 0
When x = -1, f(x) = 1
which is already done.

But find the second derivative of f(x) first.
You already have the first derivative as f'(x) = 3/4x^2 - 3/4
Find the derivative again.

Answer 23

3x/2

Answer 24

Yes. f''(x) = 3x/2
Find f''(x) at x = 1 and -1

Answer 25

3/2 and -3/2

Answer 26

i'm sorry I have been doing this study guide for literally 6 hours. These are my last two questions

Answer 27

When x = 1, f''(x) = 3/2. Since f''(3/2) > 0, x = 3/2 is a minimum.
When x = -1, f''(x) = -3/2. Since f''(-3/2) < 0, x = -3/2 is a maximum.

Now we can say the btwo points we found earlier are the extrema:
(1, 0) is a minimum
(-1, 1) is a maximum.

Since the problem asks for relative extrema you can simply say the extrema are : (-1,1) and (1,0).

You can follow the same procedure for the second problem.

Answer 28

So (-1,1) and (1,0) are the relative extrema?

Answer 29

Yes. Those are the points on the curve f(x) where the function reaches a relative maximum or a minimum.

Answer 30

For the next one the derivative is 6x^2+10x-25

Answer 31

Yes. Follow the same procedure. Give the final answer and I will check with you later.

Answer 32

x = 5(-1 sqrt7) / 6, -5(1+sqrt7) / 6

Answer 33

Thats when x=0

Answer 34

Those are the roots. That is not when x = 0. Those are the x values when h'(x) = 0.
It may be better to take the second derivative right away and evaluate h''(x) at those x values. Positive means it is a minimum, negative means it is a maximum. Then you can put the x values in h(x) and find h(x).

Answer 35

For the local minimum I get 5sqrt7-5 /6 and for the local maximum I get 5sqrt7+5 /6

Answer 36

Can you please just tell the what the relative extrema is for this last question of my answers are wrong?

Answer 37

The message and reply notifications are not working properly and so unless I visit this page I won't know if you have posted a reply. Let me check your answers...

Answer 38

Doesn't look like your minimum and maximum are correct.

Convert the x values where h'(x) = 0 to decimals first and then plug it into h(x).

Answer 39

26.46 is the maximum decimal

Answer 40

-26.46 is the minimum decimal

Answer 41

First tell me what your x values are in decimal.

Answer 42

1.37, -3.04

Answer 43

I need to have this submitted in the next 15 minutes @ranga

Answer 44

Like I said before, the message and reply notifications are not working properly.

Yes those are the correct x values. Put it in h(x) and you should get: 66.02 and -19.72

The extrema are: (1.37, -19.72) and (-3.04, 66.02)