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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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math>philosophy
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x^2 - 0?
finaynay
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i meant 9 lol
math>philosophy
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lol
math>philosophy
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So what are the conditions for the MVT
finaynay
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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
math>philosophy
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\[\sqrt{x^2} = |x|\]
finaynay
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how would i do that for |x^2-9| and did i do the beginning of the problem correctly
Callisto
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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?
finaynay
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i still don't understand
Callisto
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Which part?
finaynay
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how i'm supposed to take the derivative of \[|x ^{2}-9|\]
Callisto
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Do you know that \(\sqrt{a^2} = |a| \)?
finaynay
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yes but how do i apply that
\[\sqrt{(x-9)^{2}}\] isn't the derivative
Callisto
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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 \(\sqrt{(x-9)^2}\)
finaynay
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oh okkkkkkk but is that even what i'm supposed to be doing in this problem
Callisto
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You need to find f'(x), so, I think so :|
finaynay
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for realsssss noooooooo
Callisto
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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?
finaynay
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yeah
Callisto
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So... just differentiate that guy :|
finaynay
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so the derivative is 2x
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x=-1?
anonymous
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not quite
anonymous
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on the interval \([0,2]\) you have \(|x^2-9|=9-x^2\)
finaynay
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how is it 9−x^2
anonymous
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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
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if \(-3<x<3\) we have \(x^2-9<0\) and so \(|x^2-9|=-(x^2-9)=9-x^2\)
anonymous
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in english \(x^2-9\) is negative on the interval \([0,2]\) and so the absolute value is minus the whole thing , namely \(9-x^2\)
finaynay
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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
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lets try it
anonymous
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\(f(0)=9\) and \(f(2)=5\) and so \(\frac{f(2)-f(0)}{2-0}=\frac{5-9}{2}=-4\)
finaynay
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isn't it -2
anonymous
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ooops bad arithmetic
finaynay
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oh okay lol
math>philosophy
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<3
anonymous
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yes it is \(-2\)
now we set \(-2x=-2\) and guess what? \(x=1\)
finaynay
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-2=f'(x)
finaynay
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yessss! alright i get it now
anonymous
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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