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Find the point(s) guaranteed by the mean value theorem for f(x)=x^29 over the interval [0,2]
 one year ago
 one year ago
Find the point(s) guaranteed by the mean value theorem for f(x)=x^29 over the interval [0,2]
 one year ago
 one year ago

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math>philosophyBest ResponseYou've already chosen the best response.0
So what are the conditions for the MVT
 one year ago

finaynayBest ResponseYou've already chosen the best response.0
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^29 because i dont know how to find the derivative of a absolute value
 one year ago

math>philosophyBest ResponseYou've already chosen the best response.0
\[\sqrt{x^2} = x\]
 one year ago

finaynayBest ResponseYou've already chosen the best response.0
how would i do that for x^29 and did i do the beginning of the problem correctly
 one year ago

CallistoBest ResponseYou've already chosen the best response.0
y =x^29 Let u =x^29 => \(y=\sqrt{ u^2}\) \[\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}\] Does that help?
 one year ago

finaynayBest ResponseYou've already chosen the best response.0
i still don't understand
 one year ago

finaynayBest ResponseYou've already chosen the best response.0
how i'm supposed to take the derivative of \[x ^{2}9\]
 one year ago

CallistoBest ResponseYou've already chosen the best response.0
Do you know that \(\sqrt{a^2} = a \)?
 one year ago

finaynayBest ResponseYou've already chosen the best response.0
yes but how do i apply that \[\sqrt{(x9)^{2}}\] isn't the derivative
 one year ago

CallistoBest ResponseYou've already chosen the best response.0
It's the question :P To find the derivative of x^2 9, you can find the derivative of \(\sqrt{(x^29)^2}\), which is the same as x^2 9. Btw, it should be \(\sqrt{(x^29)^2}\), not \(\sqrt{(x9)^2}\)
 one year ago

finaynayBest ResponseYou've already chosen the best response.0
oh okkkkkkk but is that even what i'm supposed to be doing in this problem
 one year ago

CallistoBest ResponseYou've already chosen the best response.0
You need to find f'(x), so, I think so :
 one year ago

finaynayBest ResponseYou've already chosen the best response.0
for realsssss noooooooo
 one year ago

CallistoBest ResponseYou've already chosen the best response.0
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 \(\frac{f(b)f(a)}{ba} = f'(c)\) In this question, the conditions are satisfied, that is f(x) = x^29 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?
 one year ago

CallistoBest ResponseYou've already chosen the best response.0
So... just differentiate that guy :
 one year ago

finaynayBest ResponseYou've already chosen the best response.0
so the derivative is 2x
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
on the interval \([0,2]\) you have \(x^29=9x^2\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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 \(x=1\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
if \(3<x<3\) we have \(x^29<0\) and so \(x^29=(x^29)=9x^2\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
in english \(x^29\) is negative on the interval \([0,2]\) and so the absolute value is minus the whole thing , namely \(9x^2\)
 one year ago

finaynayBest ResponseYou've already chosen the best response.0
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?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
\(f(0)=9\) and \(f(2)=5\) and so \(\frac{f(2)f(0)}{20}=\frac{59}{2}=4\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
ooops bad arithmetic
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
yes it is \(2\) now we set \(2x=2\) and guess what? \(x=1\)
 one year ago

finaynayBest ResponseYou've already chosen the best response.0
yessss! alright i get it now
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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
 one year ago
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