Asked by Ivan
Consider the function f ( x ) = 3 − 5 x^2 on the interval [ − 2 , 8 ] . Find the average or mean slope of the function on this interval, i.e.
f(8)-f(-2)/8-(-2)=
By the Mean Value Theorem, we know there exists a c in the open interval ( − 2 , 8 ) such that f ' ( c ) is equal to this mean slope. For this problem, there is only one c that works. Find it.
Answer with explanation please and thanks!
f(8)-f(-2)/8-(-2)=
By the Mean Value Theorem, we know there exists a c in the open interval ( − 2 , 8 ) such that f ' ( c ) is equal to this mean slope. For this problem, there is only one c that works. Find it.
Answer with explanation please and thanks!
Answers
Answered by
oobleck
really? They gave you the actual formula.
Just evaluate m = (f(8)-f(-2))/(8-(-2)) = ___
now, since f'(x) = -10x, you want to find c such that
-10c = m
c = -m/10
Just evaluate m = (f(8)-f(-2))/(8-(-2)) = ___
now, since f'(x) = -10x, you want to find c such that
-10c = m
c = -m/10
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