rolle's + MVT?

Question

f(x) = x^2 + 1, [0,2]

for MVT...i got x=1
for Rolle's..i got x=0

is that right?

Answer 1

if you are solving for the value of c, then yes you are correct

for roll's theorem to find the value of c you set f'(x) = 0
2x = 0
x=0

for mvt to find the value of c you find the value of f(b) - f(a)/b-a
f(2) - f(0)/2-0 = (2^2 + 1) - 1/2 = 2
then take the derivative and set it equal to that solution
2x=2
x=1

Answer 2

You're right about the x fullfilling the mean value theorem, but I don't see where Rolle's theorem even applies: f(0)≠f(2), and there aren't any distinct points a, b on the interval [0, 2] such that f(a)=f(b).