Question

What is the value of c that satisfies the Mean Value Theorem (MVT) for f(x)=x^3-3x^2+x on [0, 3]?

Answer 1

f'(c) = 1/3[f(3) -f(0)]

f'(c) = 1

**f'(x) = 3x^2-6x+1 **

3c^2 -6c +1 = 1

3c^2 = 6c

c^2 = 2c

c = 0,2, since 0 is one of the endpoints, then c=2