yes. the mean value theorem lets you pick any interval, but Rolle's theorem says f(a) = f(b). It's just a special case of the MVT where f'(c) = 0. Your interval here meets that condition.
As for your exact value, forget about decimal approximations.
(-5±√70)/3 is exact. f'(1.25) is not zero.
i am on "rolles and the mean value theorem" and was just wondering, when i am doing rolles, do i really need to find the exact value of x where f'(c) = 0?
for example:
f(x) = (x+4)^2 (x-3) on [-4,3]
i get to:
3x^2+10x-8=7
then i don't know if i then need to find the exact value of x or can i say:
1.2<x<1.3
3 answers
how did you get (-5 +/- sqrt(70))/3
I assumed you wanted the roots of your equation
3x^2+10x-8=7
3x^2+10x-8=7
4 answers