Question

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?

Answer 1

`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`