Please help!! calc ab
If f(x) is differentiable for the closed interval [−1, 4] such that f(−1) = −3 and f(4) = 12, then there exists a value c, −1< c < 4 such that
f '(c) = 3
f '(c) = 0
f(c) = −15
f (c) = 3
---------
If f(x) = ∣(x2 − 10)(x2 + 1)∣, how many numbers in the interval 0 ≤ x ≤ 2.5 satisfy the conclusion of the mean value theorem?
None
One
Two
Three

- tmagloire1

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- freckles

have you tried seeing if you can apply mean value theorem to question 1 ?

- tkhunny

Mean Value Theorem... What is the slope of the extreme line segment so defined?
f(−1) = −3 and f(4) = 12

- tmagloire1

I understand that i tis mean value theorem but i didn't get the same answer as them

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## More answers

- tmagloire1

ill try it again 1 sec

- tmagloire1

The slope is 3 @tkhunny

- freckles

that is right

- freckles

so question 2
hint:
x^2+1>0 for all x
x^2-10<0 for x between -sqrt(10) and sqrt(10)
so |x^2-10|=-(x^2-10) on the interval [0,2.5]
so since you are only looking at [0,2.5]
the part of the function you only need to look at is:
\[f(x)=-(x^2-10)(x^2+1)\]

- tmagloire1

I'm totally confused. I'm sorry

- freckles

on what to do next? or what I said?

- freckles

\[|(x^2-10)(x^2+1)|=|x^2-10| \cdot |x^2+1|=|x^2-10| \cdot (x^2+1) \text{ for all } x \\ \text{ since } x^2+1 \text{ is positive for all } x \\ \\ \text{ now let } h(x)=x^2-10\]
we see that graph looks like:
|dw:1441063542991:dw|
notice the part that falls under the x-axis...
we need to reflect that little part to find the graph of p(x)=|x^2-10|
|dw:1441063593897:dw|

- freckles

so
\[|x^2-10|=x^2-10 \text{ on } (-\infty,-\sqrt{10}] \cup [\sqrt{10},\infty) \\ \text{ or } \\ |x^2-10|=-(x^2-10) \text{ on } (-\sqrt{10},\sqrt{10})\]

- freckles

so on [0,2.5] our function looks like
\[f(x)=-(x^2-10)(x^2+1)\]

- freckles

now apply mean value theorem

- freckles

\[f'(c)=\frac{f(2.5)-f(0)}{2.5-0}\]
solve for c
you will need first find f'(x)
and then find f'(c) by replacing x with c
you will also of course need to evaluate f(2.5) and f(0)

- freckles

and then finally solve the equation for c

- tmagloire1

I got 17.18/2.5

- freckles

hmm... I'm getting 17.1875 on the top
maybe I did something wrong
or did you approximate?

- freckles

\[f'(c)=\frac{17.1875}{2.5} \\ f'(c)=6.875\]
ok but you still need to work with the left hand side of the equation

- freckles

you need to find f'

- tmagloire1

Wait so if f'(c) =6.875 what would use evaluate for the left side? @freckles

- freckles

the function f
but you have to find derivative of f to find f'

- freckles

have you differentiated the function on [0,2.5] yet!

- tmagloire1

18x-4x^3

- freckles

\[f'(c)=6.875 \\ 18c-4c^3=6.875\]

- tmagloire1

ohh ok and just solve for c

- freckles

\[-4c^3+18c-6.875=0\]
well you can solve for c
but you only need to state the amount of solutions to this equation on [0,2.5]

- tmagloire1

I got 2 roots

- freckles

so you could graph -4x^3+18x-6.875 on your calculator
and see how many times it crosses the x-axis on [0,2.5]
and yeah should be 2
that is what i see too

- tmagloire1

one last thing, for the first problem. was the answer 3 for a differentiated function or for the normal answer.

- freckles

you do know the mean-value theorem if it satisfied the conditions then there c is between [-1,4] such that
\[f'(c)=\frac{f(4)-f(-1)}{4-(-1)}\]
do you know notice it says f'(c) and not f(c)?

- tmagloire1

Oh okay I missed that. Thank you for helping me with these problems! Sorry my internet was really slow and i lost connecivity half way through

- freckles

it's cool

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