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anonymous
 5 years ago
if f(x)=abs((x^212)(x^2+4)), how many numbers in the interval 2<x<3 satisfy the conclusion of the mean value theorem?
thank you so much and please tell me how u got the answer.
anonymous
 5 years ago
if f(x)=abs((x^212)(x^2+4)), how many numbers in the interval 2<x<3 satisfy the conclusion of the mean value theorem? thank you so much and please tell me how u got the answer.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0First, you need to put this function into something you can handle. You can use the definition of absolute value to do this:\[f(x)=(x^212)(x^2+4)=\sqrt{[(x^212)(x^2+4)]^2}\]so\[f(x)=\sqrt{(x^212)^2(x^2+4)^2}\]\[=\sqrt{x^816 x^632 x^4+768 x^2+2304}\]\[=\sqrt{(x^48 x^248)^2}\]Now, the mean value theorem says that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that\[f'(c)=\frac{f(b)f(a)}{ba}\]You need to find all those c's in (2,3) that satisfy,\[f'(c)=\frac{f(3)f(2)}{32}=f(3)f(2)\]...end part 1

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0only that its (2, 3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol I love how you always clean up the question.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nowhereman, do you want to finish this? I want to go eat.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, so you are looking for the points \[c \in (2, 3)\] so that \[f'(c) = \frac{f(3)  f(2)}{3(2)} = \frac{3964}{5} = 5\] So we need to find the solutions of \[\frac{2(x^4  8x^248)(4x^3  16x)}{2\sqrt{(x^48x^248)^2}} = 5\]

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Oh and we can do it a little bit easier, because \[x \in [2, 3] ⇒ x^212 < 0 ∧ x^2+4 > 0 \]\[⇒ f(x) = (x^212)(x^2+4) = (x^212)(x^2+4)\]

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0So \[f'(x) = 4x^3 + 16x\] and we need to solve \[4x^3 + 16x  5 = 0\] by differentiating you get that the extrema of \[4x^3 + 16x  5\] are at \[\pm 2\sqrt{\frac{1}{3}}\]

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0the function values are there \[\pm32\frac{2}{3}\sqrt{\frac 1 3}5\] one is bigger then zero and the other is smaller then zero. And so there must be 3 different solutions.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0you should maybe draw the graph of f' or let some plotter do it for you, to get a better imagination of the situation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you both so much for providing such a detailed answer! u have saved my calc grade!
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