the average rate of change of f on [a,b] is just [f(b)-f(a)]/(b-a)
f(x) = x^3-x^2+4x+c
The c does not matter, so the average rate of change on [-1,4] is just
[f(4)-f(-1))/3 = ((64-16+16+c)-(-1-1-4+c))/3 = 70/3
You want a c where f'(c) = 70/3
3c^2-2c+4 = 70/3
c = (1+√59)/3 = 2.89
which is in [-1,4]
since you know f(-1), you have
f(-1) = -1-1-4+c = -2
c = 4
as always, double-check my math - this was done on the fly.
Suppose f"(x)=6x-2
A. If f'(1)=5, find f'(x).
I got f'(x)=3x^2-2x+4
B. Fine the average rate of change of f on [-1,4].
C. Fine the value of x guaranteed by the mean value theorem for r on [-1,4].
D. If f(-1)=-2, find f(x)
I am not sure how to do B, C, and D.
1 answer
1 answer