Given the function #f(x)=x(x^2-x-2)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [-1,1] and find the c?

1 Answer
Feb 13, 2018

#c = -1/3#

Explanation:

first of all, the Mean Value theorem isn't an hypothesis, it is a theorem which states that if a function is continuous over an interval #[a,b]# and Differentiable on the interval #(a,b)# then there exists a value c such that the slope of the tangent line at c is equal to the overall slope from a to b and #a < c < b#

and since this function is continuous over all values
, we know that there does exist such a value c, all we have to do is find it

#f'(c) = (f(b) - f(a))/(b-a)#

where #a = -1 and b = 1#

therefore
#f'(c) = -1#

the derivative of this function can be found fairly simply by just using the power rule and the derivative is
#f'(x) = 3x^2 - 2x - 2#

therefore, to find c,
#3c^2 - 2c - 2 = -1#
#3c^2 - 2c - 1 = 0#

and on solving this equation using the quadratic formula
we Get #c = -1/3 or 1#

but since c should be such that #a < c < b #, that is, c should be in the middle of the endpoints, #c = 1# cannot be an answer as 1 is one of the endpoint.

Therefore
#c = -1/3#