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Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
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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
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 \(\sqrt{(x-9)^2}\)
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?
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