- anonymous

Graph the function f(x)=x^3−2x and it's secant line through the points (-2,-4) and (2,4). Use the graph to estimate the x-coordinate of the points where the tangent line is parallel to the secant line.
Find the exact value of the numbers c that satisfy the conclusion of the mean value theorem for the interval [-2,2].

- schrodinger

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- anonymous

I need help finding 'c'

- anonymous

I know that the Mean Value Theorem says... \[\frac{ f(b)-f(a) }{ b-a }=f'(c)\]

- anonymous

just apply that...

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## More answers

- anonymous

But that gives me f'(c)

- anonymous

lemme check

- anonymous

here a=-2
b=2

- anonymous

Right, plugging all that in do you get f'(c) = 2 ?

- anonymous

i get 8

- anonymous

How?

- anonymous

f(b)=2^3-2(2)..rite?

- anonymous

f(a)=(-2)^3-2(-2)

- anonymous

Doesn't it already give us the coordinates for the points? (2,4) and (-2,-4)

- anonymous

Is it wrong to say \[\frac{ 4+4 }{ 2+2}=\frac{ 8 }{ 4 } = 2\] ? Or am I totally wrong right now? Lol

- anonymous

what did u do just now?

- myininaya

\[(x^3-2x)'|_{x=c}=\frac{f(2)-f(-2)}{2-(-2)} \\ (x^3-2x)'|_{x=c}=\frac{4-(-4)}{2-(-2)} \\ (x^3-2x)'|_{x=c}=\frac{4+4}{2+2}\]
that is right for the right hand side

- anonymous

Well I plugged in f(a) = -4, f(b) = 4, and a=-2, b=2 in the mean value theorem formula

- anonymous

what happens to the left hand side?

- anonymous

do i just derive it?

- myininaya

you still have to differentiate the left hand side and plug in c afterwards as the equation above suggests

- anonymous

3x^2 - 2 right?

- myininaya

yep and plug in c (though that part doesn't matter to much; it is just what they call it in the formula above)
yep now solve 3x^2-2=2

- myininaya

or you named it c
3c^2-2=2

- anonymous

Ohh so then its +/- 2/sqrt3 right?

- myininaya

are both of those in the given interval?

- anonymous

hmmm... yes?

- myininaya

ok then cool stuff

- anonymous

Thats it? :O

- myininaya

that is it

- myininaya

though there are two problems

- myininaya

did you do the first problem?

- anonymous

Yay!!! Thats correct! :D Thank you so much!

- myininaya

we did the second problem
we found the exact solutions

- anonymous

Yeah Im always better in graphing haha

- myininaya

alright

- anonymous

Thanks again for the help! :)

- myininaya

np

- anonymous

tracy?got the answer for the second part?

- anonymous

Yeah! I got it, its + or - 2/sqrt3

- anonymous

Thank you too @chethus!

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