What values (c) if any are predicted by the mean value theorem for the function f(x)= (x-2)^3 on the interval [0,2]?
I got x= 4 and x=0, since x=0 is within the interval I chose that as my answer. Thank you.
3 answers
Sorry I forgot to include the topic.
MVT says that there is some c in [0,2] where f'(c) = [f(2)-f(0)]/2
f(2) = 0
f(0) = -8
[f(2)-f(0)]/2 = 8/2 = 4
MVT does not predict values. It just says that at some c in [0,2] f'(c) = 4.
Now, we can check that by noting
f'(x) = 3(x-2)^2, so
f'(0) = 8 and f'(2) = 0, so the slope decreases from 8 to 0 in the interval. Since f' is continuous, f'=4 somewhere on the interval.
3(x-2)^2 = 4
x-2 = ±2/√3
x = 2-2/√3 = 0.85
f(2) = 0
f(0) = -8
[f(2)-f(0)]/2 = 8/2 = 4
MVT does not predict values. It just says that at some c in [0,2] f'(c) = 4.
Now, we can check that by noting
f'(x) = 3(x-2)^2, so
f'(0) = 8 and f'(2) = 0, so the slope decreases from 8 to 0 in the interval. Since f' is continuous, f'=4 somewhere on the interval.
3(x-2)^2 = 4
x-2 = ±2/√3
x = 2-2/√3 = 0.85
Thank you.