How many values of c satisfy the conclusion of the Mean Value Theorem for f(x) = x^3 + 1 on the interval [-1,1]?

1 Answer
Bill K.
Jun 17, 2015

There are two values of #c# that work.

Explanation:

The derivative of #f# is #f'(x)=3x^2#. The slope of the secant line is #(f(1)-f(-1))/(1-(-1))=(2-0)/2=1#. Setting #f'(x)=1# and solving for #x# results in #x=\pm 1/sqrt{3}#. These are the two "#c#" values, and they are in the interval #[-1,1]#.

Here's a picture for the given situation:

enter image source here

The red line is the secant line between #(-1,f(-1))=(-1,0)# and #(1,f(1))=(1,2)#. The green line is the tangent line to #f# at #(-1/sqrt(3),f(-1/sqrt(3)))\approx (-0.577,0.808)# and the gold line is the tangent line to #f# at #(1/sqrt(3),f(1/sqrt(3)))\approx (0.577,1.192)#. All these lines are parallel with a slope of 1.

Related questions
See all questions in Mean Value Theorem for Continuous Functions
Impact of this question
12074 views around the world
You can reuse this answer
Creative Commons License