How do you find the number c that satisfies the conclusion of the Mean Value Theorem for the function #f(x) = x^3 + x^2# on the interval [0,1]?

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Answer 1 · 2016-08-24T18:03:42

Find all solutions to #f'(x) = (f(1)-f(0))/(1-0)#. The #c#'s of the MVT conclusion must be in #(0,1)#

#f'(x) = 3x^2+2x#

#f(1) = 1^3 + 1^2 = 1+1 = 2#

#f(0) = 0^3+0^2 = 0+0 = 0#

So you need to solve

#3x^2 + 2x = (2-0)/(1-0) = 2# which is equivalent to

#3x^2+2x-2 = 0#.

This won't factor easily, so use the quadratic formula or complete the square to get

#x = (-2 +-sqrt(2^2 - 4(3)(-2)))/(2(3))#

# = (-2 +- sqrt(4+24))/6#

# = (-2 +- sqrt(28))/6#

# = (-2+- 2 sqrt7)/6#

# = (-1+-sqrt7)/3#

Sinced #(-1-sqrt7)/3 < 0#, it cannot be a #c# for the conclusion of the mean value theorem on the interval #[0,1]#.

#(-1+sqrt7)/3# is in the interval #(0,1)#, so it will work as a #c# in his question.

We know that #(-1+sqrt7)/3# is in the interval #(0,1)# because

#2 < sqrt 7 < 3# so

#-1+2 < -1 + sqrt7 < -1 + 3# and

# 1/3 < (-1+sqrt7)/3 < 2/3#.