Question

Determine whether the Mean Value Theorem can be applied to f on the closed interval [a, b]. If the Mean Value Theorem can be applied, find all values of c in the open interval (a, b) such that f '(c) = f (b) − f (a) b − a . If the Mean Value Theorem cannot be applied, explain why not.

f(x) = x^3 + 2x, [-1,1]

Answer 1

There is a value of
`c` in (-1, 1),
`c = 0.577`.

Explanation:

Let
`f(x) = x^(3)+2\cdot x` for
`x \in[-1,1]`, we need to prove that
`f(x)` is continuous and differentiable to apply the Mean Value Theorem. Given that
`f(x)` is a polynomical function, its domain comprises all real numbers and therefore, function is continuous.

If
`f(x)` is differentiable, then
`f'(x)` exists for all value of
`x`. By definition of derivative, we obtain the following expression:

`f'(x) = \lim_(h \to 0) (f(x+h)-f(x))/(h)`

`f'(x) = \lim_(h \to 0) ((x+h)^(3)+2\cdot (x+h)-x^(3)-2\cdot x)/(h)`

`f'(x) = \lim_(h \to 0) (x^(3)+3\cdot x^(2)\cdot h+3\cdot x\cdot h^(2)+h^(3)+2\cdot x+2\cdot h-x^(3)-2\cdot x)/(h)`

`f'(x) = \lim_(h \to 0) (3\cdot x^(2)\cdot h+3\cdot x\cdot h^(2)+h^(3)+2\cdot h)/(h)`

`f'(x) = \lim_(h \to 0) 3\cdot x^(2)+ \lim_(h \to 0) 3\cdot x \cdot h+ \lim_(h \to 0) h^(2)+ \lim_(h \to 0) 2`

`f'(x) = 3\cdot x^(2)+2` (Eq. 2)

The derivative of a cubic function is quadratic function, which is also a polynomic function. Hence, the function is differentiable at the given interval.

According to the Mean Value Theorem, the following relationship is fulfilled:

`f'(c) = (f(1)-f(-1))/(1-(-1))` (Eq. 3)

If we know that
`f(-1) = -3`,
`f(1) = 3` and
`f'(c) = 3\cdot c^(2)+2`, then we expand the definition as follows:

`3\cdot c^(2)+2 = 3`

`3\cdot c^(2) = 1`

`c = \sqrt{(1)/(3) }`

`c \approx 0.577`

There is a value of
`c` in the interval (-1, 1),
`c = 0.577`.