Question

Show `|f(b)-f(a)|<= |b-a|` ?

Given `f:RR->RR` differentiable in `RR` with `f'(x)=(2x)/(x^2+1)` , `AAx` `in` `RR`

Show that `|f(b)-f(a)|<= |b-a|`

`a,b` `in` `RR`

Note: Don't try to find `f` - (no given values for constant.)

Answer 1

We have:

`f'(x) = (2x)/(x^2+1) \ \ AA x in RR`

By the Mean Value Theorem, `EE c in RR` such that:

`f(b)-f(a) = f'(c) (b-a)`

Hence, we have:

`|f(b)-f(a)| = |f'(c) (b-a) | = |f'(c)| \ |(b-a) |`

So we can reduce the problem to that of proving that:

`|f'(c)| le 1`

Which requires us to consider the maximum value of `f'(x)`, by finding its critical points, using its derivative, f''(x). So differentiating wrt `x`, by applying the quotient rule:

`f''(x) = { (x^2+1)(2) - (2x)(2x) ) / (x^2+1)^2`
`\ \ \ \ \ \ \ \ \ \ = (2x^2+2-4x^2) / (x^2+1)^2`
`\ \ \ \ \ \ \ \ \ \ = (2-2x^2) / (x^2+1)^2`

So, at a critical point of `f'(x)`, we require its derivative to vanish:

`:. (2-2x^2) / (x^2+1)^2 = 0 => x^2-1=0 => x=+-1`

When:

`{ (x=-1,=>f'(x) = -1), (x=1,=>f'(x) = 1) :}`

To determine the nature of the critical points, we could consider the second derivative of `f'(x)`, ie `f'''(x)`, but equally we can examine a graph of ten function:
graph{y=(2x)/(x^2+1) [-6.24, 6.25, -3.12, 3.12]}

And we can see that:

`{ ((-1,1),=> \ "minimum"), ((1,1),=> \ "maximum") :}`

As such we can conclude that:

The range of `f'(x)` is `-1 le f'(x) le 1 => |f'(x)| le 1`

Then returning to the original problem, we can conclude that

`|f(b)-f(a)| le |(b-a) | \ \ \ ` QED

Answer 2

As was explained in Steve M's answer in https://socratic.org/s/aQwiERBL , we can prove the result using the mean value theorem, provided we can prove that `|f^'(x)| <1` `forall x in RR`.

Explanation:

We provide an algebraic proof of this below:

`f^'(x) = (2x)/(x^2+1) implies`

`f^'(x) + 1 = (2x+x^2+1)/(x^2+1) = (1+x)^2/(x^2+1)>=0quad forall x in RR`

and

`1-f^'(x) = (x^2+1-2x)/(x^2+1) = (x-1)^2/(x^2+1) >= 0quad forall x in RR`

Thus `forall x in RR`, we have

`-1 <= f^'(x) <= 1qquad implies qquad |f^'(x)| <= 1`

The rest of the proof proceeds as in https://socratic.org/s/aQwiERBL