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.

Answer 1

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