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

## Given $f : \mathbb{R} \to \mathbb{R}$ differentiable in $\mathbb{R}$ with $f ' \left(x\right) = \frac{2 x}{{x}^{2} + 1}$ , $\forall x$$\in$$\mathbb{R}$ Show that $| f \left(b\right) - f \left(a\right) | \le | b - a |$ $a , b$$\in$$\mathbb{R}$ Note: Don't try to find $f$ - (no given values for constant.)

Apr 30, 2018

We have:

$f ' \left(x\right) = \frac{2 x}{{x}^{2} + 1} \setminus \setminus \forall x \in \mathbb{R}$

By the Mean Value Theorem, $\exists c \in \mathbb{R}$ such that:

$f \left(b\right) - f \left(a\right) = f ' \left(c\right) \left(b - a\right)$

Hence, we have:

$| f \left(b\right) - f \left(a\right) | = | f ' \left(c\right) \left(b - a\right) | = | f ' \left(c\right) | \setminus | \left(b - a\right) |$

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

$| f ' \left(c\right) | \le 1$

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

$f ' ' \left(x\right) = \frac{\left({x}^{2} + 1\right) \left(2\right) - \left(2 x\right) \left(2 x\right)}{{x}^{2} + 1} ^ 2$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{2 {x}^{2} + 2 - 4 {x}^{2}}{{x}^{2} + 1} ^ 2$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{2 - 2 {x}^{2}}{{x}^{2} + 1} ^ 2$

So, at a critical point of $f ' \left(x\right)$, we require its derivative to vanish:

$\therefore \frac{2 - 2 {x}^{2}}{{x}^{2} + 1} ^ 2 = 0 \implies {x}^{2} - 1 = 0 \implies x = \pm 1$

When:

$\left\{\begin{matrix}x = - 1 & \implies f ' \left(x\right) = - 1 \\ x = 1 & \implies f ' \left(x\right) = 1\end{matrix}\right.$

To determine the nature of the critical points, we could consider the second derivative of $f ' \left(x\right)$, ie $f ' ' ' \left(x\right)$, 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:

$\left\{\begin{matrix}\left(- 1 1\right) & \implies \setminus \text{minimum" \\ (11) & => \ "maximum}\end{matrix}\right.$

As such we can conclude that:

The range of $f ' \left(x\right)$ is $- 1 \le f ' \left(x\right) \le 1 \implies | f ' \left(x\right) | \le 1$

Then returning to the original problem, we can conclude that

$| f \left(b\right) - f \left(a\right) | \le | \left(b - a\right) | \setminus \setminus \setminus$ QED

Apr 30, 2018

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 $| {f}^{'} \left(x\right) | < 1$ $\forall x \in \mathbb{R}$.

#### Explanation:

We provide an algebraic proof of this below:

${f}^{'} \left(x\right) = \frac{2 x}{{x}^{2} + 1} \implies$

${f}^{'} \left(x\right) + 1 = \frac{2 x + {x}^{2} + 1}{{x}^{2} + 1} = {\left(1 + x\right)}^{2} / \left({x}^{2} + 1\right) \ge 0 \quad \forall x \in \mathbb{R}$

and

$1 - {f}^{'} \left(x\right) = \frac{{x}^{2} + 1 - 2 x}{{x}^{2} + 1} = {\left(x - 1\right)}^{2} / \left({x}^{2} + 1\right) \ge 0 \quad \forall x \in \mathbb{R}$

Thus $\forall x \in \mathbb{R}$, we have

$- 1 \le {f}^{'} \left(x\right) \le 1 q \quad \implies q \quad | {f}^{'} \left(x\right) | \le 1$

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