Find the point(s) guaranteed by the mean value theorem for f(x)=|x^2-9| over the interval [0,2]

- anonymous

Find the point(s) guaranteed by the mean value theorem for f(x)=|x^2-9| over the interval [0,2]

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- anonymous

x^2 - 0?

- anonymous

i meant 9 lol

- anonymous

lol

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## More answers

- anonymous

So what are the conditions for the MVT

- anonymous

well it is already guaranteed that the mean value theorem applies so i did f'(x)= f(2)-f(0)/2
f'(x)=-2
then what do i find the derivative of f(x)=|x^2-9| because i dont know how to find the derivative of a absolute value

- anonymous

\[\sqrt{x^2} = |x|\]

- anonymous

how would i do that for |x^2-9| and did i do the beginning of the problem correctly

- Callisto

y =|x^2-9|
Let u =x^2-9
=> \(y=\sqrt{ u^2}\)
\[\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}\]
Does that help?

- anonymous

i still don't understand

- Callisto

Which part?

- anonymous

how i'm supposed to take the derivative of \[|x ^{2}-9|\]

- Callisto

Do you know that \(\sqrt{a^2} = |a| \)?

- anonymous

yes but how do i apply that
\[\sqrt{(x-9)^{2}}\] isn't the derivative

- Callisto

It's the question :P
To find the derivative of |x^2 -9|, you can find the derivative of \(\sqrt{(x^2-9)^2}\), which is the same as |x^2 -9|.
Btw, it should be \(\sqrt{(x^2-9)^2}\), not \(\sqrt{(x-9)^2}\)

- anonymous

oh okkkkkkk but is that even what i'm supposed to be doing in this problem

- Callisto

You need to find f'(x), so, I think so :|

- anonymous

for realsssss noooooooo

- Callisto

Hmmm.. MVT:
Suppose y=f(x) is continuous on a closed interval [a,b] and differentiable on the interior (a,b). Then there is at least one point c at which \(\frac{f(b)-f(a)}{b-a} = f'(c)\)
In this question, the conditions are satisfied, that is f(x) = |x^2-9| is continuous on the closed interval [0, 2] and is differentiable on (0, 2)
So, just apply it: \(\frac{f(2)-f(0)}{2} = f'(c)\)
And you have to find f'(x), isn't it?

- anonymous

yeah

- Callisto

So... just differentiate that guy :|

- anonymous

so the derivative is 2x

- anonymous

x=-1?

- anonymous

not quite

- anonymous

on the interval \([0,2]\) you have \(|x^2-9|=9-x^2\)

- anonymous

how is it 9−x^2

- anonymous

since this is a quadratic, the "point guaranteed to exist by the mean values theorem will be in the midpoint of the interval, namely at \(x=1\)

- anonymous

if \(-3

- anonymous

in english \(x^2-9\) is negative on the interval \([0,2]\) and so the absolute value is minus the whole thing , namely \(9-x^2\)

- anonymous

okay so the derivative is -2x right and then you set that equal to the mean and you solve for x and you get x=1?

- anonymous

lets try it

- anonymous

\(f(0)=9\) and \(f(2)=5\) and so \(\frac{f(2)-f(0)}{2-0}=\frac{5-9}{2}=-4\)

- anonymous

isn't it -2

- anonymous

ooops bad arithmetic

- anonymous

oh okay lol

- anonymous

<3

- anonymous

yes it is \(-2\)
now we set \(-2x=-2\) and guess what? \(x=1\)

- anonymous

-2=f'(x)

- anonymous

yessss! alright i get it now

- anonymous

you can convince yourself, if you have the time in inclination, that the "point guaranteed to exist by the mean value theorem" is smack in the middle of the interval if your function is a quadratic

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