Question

How do you determine whether the function `f(x) = abs(x-3)` satisfies the hypotheses of the mean value theorem on the indicated interval (a,b), and if so how do you find all numbers c on [0,4] ?

Answer 1

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)`.

In this question, `f(x) = abs(x-3)` , `a=0` and `b=4`.

We can apply the Mean Value Theorem if both hypotheses are tru
This function is continuous on its domain, `9-oo, oo)`, so it is continuous on `[0, 4]`

Now,
`f(x) =abs(x-3) = { (-x+3, ", if x < 3"), ( x-3, ", if x > 3") :}`

So, for the derivative, we have:

`f'(x) = { (-1, ", if x < 3"), ( 1, ", if x > 3") :}`

So `f` is not differentiable at `3`, so it is not differentiable on `(0, 0)`

This function does not satisfy the hypotheses on `[0,4]`.

In addition , since the derivative is always `-1` or `1` (where it exists),

and `(f(4)-f(0))/(4-0) = 1/2`

There is no `c in (0,4)` at which `f'(c) = (f(4)-f(0))/(4-0)`.

(There is no `c` that satisfies the conclusion.)