Question

Calc Help? Check my work

Answer 1

I found the derivative of - x^2-1/(x^2+1)^2. I then set it to zero to get x=1,-1 as our critical points. I then plugged in -2 and 2 into the derivative to see whether it has a negative or positive slope. The relative extrema is x=0 is the local max and x=sqrt3,-sqrt3 is a local minimum.

Answer 2

Your derivative looks a little off,
unless you made a typo,\[\large\rm f'(x)=\frac{1-x^2}{(x^2+1)^2}\]the 1 should be positive.

Answer 3

Critical points look correct though :) Hmm

Answer 4

in my calc it shows the whole function to be negative the negative is before the equation...

Answer 5

Does it? Ok maybe I made a boo boo :) I'll check again

Answer 6

That's ok once you distribute you get what zep has

Answer 7

Oh did you mean to write this?\[\large\rm f'(x)=-\frac{x^2-1}{(x^2+1)^2}\]

Answer 8

I thought the negative was on the x^2 only, my bad.

Answer 9

yes that is what i got

Answer 10

I feel like you're missing something for your test points.
You wanted to see what was happening "around" each test point.

So you chose x=-2 to see what is happening on the left side of x=-1, good.
You then chose x=2 to see what is happening on the right side of x=2, good.
What about choosing a test point between x=-1 and x=1?

Answer 11

to see what happening on the right side of x=1*

Answer 12

ill plug one in give me a sec

Answer 13

if we plug in 0 we get 1 so it is a positive slope

Answer 14

ok sounds good :)

Answer 15

so then my answer is correct i just have to add in that i tested 0?

Answer 16

x=0 was not a critical point.
How could it possibly be extrema? 0_O

Answer 17

it is the local max

Answer 18

??

Answer 19

You get your possible max/min values from your critical points.
x=0 is not a critical point, so it's not even in the running.

Answer 20

but when we set it to 0 we get our critical points?

Answer 21

This answer key makes no sense at all...
where is sqrt(3) coming from?

Are you sure this corresponds to this problem?

Answer 22

that is when i plug the derivative into a calc and it says find the local min and max

Answer 23

Oh oh oh, this is a different problem.
See that f(x) does not match your f(x).

Answer 24

oh should i have used f'(x) or the original f(x)

Answer 25

You used f'(x), they wanted f(x) apparently.
But I hope what I'm explaining makes sense without you needing the calculator lol :)

x=0 was not a critical point,
x=-sqrt(3) doesn't even make sense XD lol

Answer 26

Haha, yeah so essentially what you do is find the derivative and set it equal to 0 and find where it's undefined to find the candidates for your critical values.

Answer 27

No no, he used f(x) he didn't find f'(x)

Answer 28

No, I'm pretty sure his original post includes his f(x),
this newer post is some online calculator.

Answer 29

Haha xD

Answer 30

yeah i found the derivative and set it to 0 to find the critical points... the original f(x) is the other thing the one i posted was the dervative not the original f(x)

Answer 31

ok so where do i take it from here?

Answer 32

Ok I see you had it listed as f(x)

Answer 33

yeah it should have been f'(x)

Answer 34

you applied the first derivative test, did your sign chart good.
Do you see your local max min?

Answer 35

-1 being the min adn 1 being the max?

Answer 36

ya :d simple as that :D

Answer 37

thanks i will post the second question in a second

Answer 38

Err careful how you say that :)
They usually want the y coordinate for the max/min.

So we would say x=-1 gives us a minimum value of y=-1/2

Answer 39

ok

Answer 40

here is the second question it uses the MVT

Answer 41

yeah

Answer 42

For us to be allowed to apply the MVT, the function must satisfy two things:
~ Continuous on this interval [0,1]
~ Differentiable on this interval (0,1)

Do we have any discontinuities?
Does this function have any asymptotes or sharp corners?

Answer 43

Vertical Asymptotes :x=−1
Horizontal Asymptotes :y=0

Answer 44

x=-1 is a discontinuity

Answer 45

Discontinuity at x=-1 which is NOT within our interval [0,1].
If we take the derivative of this function, we'll end up with some type of x+1 in the denominator, which still only tells us about x=-1.

So good, our function is continuous and differentiable on [0,1],
so yes we can apply the MVT :)
Step 1 complete.

Answer 46

ok

Answer 47

If we take the secant line which connects the ends points of our interval,