Question

Given the function `f(x) = 8 sqrt{ x} + 1`, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [1,10] and find the c?

Answer 1

Please see below.

Explanation:

The Mean Value Theorem has two hypotheses:

H1 : `f` is continuous on the closed interval `[a,b]`

H2 : `f` is differentiable on the open interval `(a,b)`.

A hypothesis is satisfied if it is true.

So we need to determine whether

`f(x) = 8sqrtx+1` is continuous on the closed interval `[1,10]`.
It is.

(`f` is continuous on its domain, `[0,oo)`, so it is continuous on `[1,10]`.)

and whether

`f(x) = 8sqrtx+1` is differentiable on the open interval `(1,10)`.
It is.

(Remember that "differentiable" means the derivative exists.)
`f'(x) = 4/sqrtx` exists for all `x > 0`. So, `f` is differentiable on `(1,10)`.)

A very common way of stating the Mean Value Theorem gives the conclusion as:

There is a `c` in `(a,b)` such that `f'(c) = (f(b)-f(a))/(b-a)`

A popular exercise (calculus and algebra) is to ask students to find the value or values of `c` that satisfy this conclusion.

To do this, find `f'(x)`, do the arithmetic to find `(f(b)-f(a))/(b-a)` and solve the equation

`f'(x) = (f(b)-f(a))/(b-a)` on the interval (a,b)#