anonymous
  • anonymous
Find the point(s) guaranteed by the mean value theorem for f(x)=|x^2-9| over the interval [0,2]
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
x^2 - 0?
anonymous
  • anonymous
i meant 9 lol
anonymous
  • anonymous
lol

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