Asked by matt
find the point, (c,f(c)), on f(x)=x^2 in the interval 0≤X≤4, such that f'(c) equals the average rate of change over that interval.
Answers
Answered by
Steve
This is just the Mean Value Theorem. Find c such that
f'(c) = (f(4)-f(0)/(4-0)
= 16/4 = 4
Since f'(x) = 2x, where does 2x=4?
y-4 = 4(x-2)
http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E2,+y%3D4(x-2)%2B4,+y%3D4x
f'(c) = (f(4)-f(0)/(4-0)
= 16/4 = 4
Since f'(x) = 2x, where does 2x=4?
y-4 = 4(x-2)
http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E2,+y%3D4(x-2)%2B4,+y%3D4x
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