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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So what are the conditions for the MVT

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how would i do that for x^29 and did i do the beginning of the problem correctly

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i still don't understand

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

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.0Do you know that \(\sqrt{a^2} = a \)?

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

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.0It'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 \(\sqrt{(x9)^2}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh okkkkkkk but is that even what i'm supposed to be doing in this problem

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.0You need to find f'(x), so, I think so :

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for realsssss noooooooo

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.0Hmmm.. 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?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.0So... just differentiate that guy :

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the derivative is 2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0on the interval \([0,2]\) you have \(x^29=9x^2\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since 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
 3 years ago
Best ResponseYou've already chosen the best response.0if \(3<x<3\) we have \(x^29<0\) and so \(x^29=(x^29)=9x^2\)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay 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
 3 years ago
Best ResponseYou've already chosen the best response.0\(f(0)=9\) and \(f(2)=5\) and so \(\frac{f(2)f(0)}{20}=\frac{59}{2}=4\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes it is \(2\) now we set \(2x=2\) and guess what? \(x=1\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yessss! alright i get it now

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you 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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