f'(x)=1/3 (x)^-2/3
f (1)=1^1/3=1
f (8) = 2
so, the point c must be such that
f'(c)= (f(8)-f(1))/(8-1)=(2-1)/7=1/7
but f'(c)=1/3 (c)^-2/3 and f'(c)=1/7 so
1/7=1/3(c^-2/3)
3/7=c^-2/3
take each side to the 3/2 power
c^2/3=7/3
c= cube root (7/3)^2 = 1.76
so, the conclusion of the theorem is borne out.
Find a point c satisfying the conclusion of the Mean Value Theorem for the function f(x)= x^1/3 on the interval [1,8]
I got f'(c)= 1/7 but am not sure where to go from there.
1 answer
1 answer