Determine if Mean Value Theorem applies to function

Question

Determine if the Mean Value Theorem for Integrals applies to the function f(x) = x³ - 9x on the interval [-1, 1]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem.

Answer 1

1. f is continuous on the closed interval [-1,1] as it is a polynomial function
2. f is differentiable on the open interval (-1,1), since the derivative f'=3x^2-9 exists and f' is continuous
3. The mean value theorem applies: There is at least one value x=c so that the slope m of the secant through the points (x=1,y=f(1) ) and ( x=-1,y=f(-1) ) is equal to the slope f'(c) of some tangent. (i.e tangent line and secant line are parallel. We will see that there are two such tangent lines). Solve

f(1)-f(-1)=f'(c)(1-(-1))----> Plug in : f(1)=-8, f(-1)=8

-8-8=(3c^2-9)*2---------> -16=6c^2-18---------> 2=6c^2----->1/3 =c^2-----> c= +or - sqrt(1/3)

There are two tangents: For c = sqrt(1/3) and for c=-sqrt(1/3) you get m=f'(1/3)=3*(1/3)^2 - 9 =1/3 -9 = -26/3 (slope is negative). Make a drawing an confirm. One tangent touches the curve at (x=1/3, y=f(1/3) ).

THe other tangent touches the curve at (x=-1/3, y=f(-1/3)).