Verify the means value theorem holds on the interval shown. Then, find the value c such that f'(c)=(f(b)-f(a))/(b-a)
a. f(x)= x-1/x on [1,3]
b.f(x)=x^3=x-4 on [-2,3]
c. f(x)= x^3 on [-1,2]
d. f(x)= Sqr. root of x on [0,4]
Well, try one of them.
function is continuous and differentiable from 1 to 3 (note not at zero)
f(3) = 3-1/3 = 8/3
f(1) = 0
[f(3)-f(1)]/(3-1) = 4/3
so where is the derivative 4/3?
f'(x) = 1 + 1/x^2
where does 1+1/x^2 = 4/3 ?
1/x^2 = 1/3
x = sqrt 3
yes, that is between x = 1 and x =3