# How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for  f(x) = sqrt(x^2+1) ?

Nov 11, 2016

If  { (x<0, => f'(x)<0, =>, "strictly decreasing"), (x=0, => f'(x)=0, =>, "stationary"), (x>0, => f'(x)>0, =>, "strictly increasing") :}

minimum at $\left(0 , 1\right)$

#### Explanation:

We have $f \left(x\right) = \sqrt{{x}^{2} + 1}$

graph{sqrt(x^2+1) [-10, 10, -5, 5]}

We can deduce from the graph that

f(x) is  { ("strictly decreasing", x<0), ("stationary", x=0), ("strictly increasing", x>0) :}

So let's prove this using calculus.

For the sake of simpler notation, Let $y = \sqrt{{x}^{2} + 1}$, and we can rearrange and differentiate implicitly. We don't need an explicit expression for $\frac{\mathrm{dy}}{\mathrm{dx}}$ we just need to identify the critical point ($\frac{\mathrm{dy}}{\mathrm{dx}} = 0$)

$y = \sqrt{{x}^{2} + 1}$
$\therefore {y}^{2} = {x}^{2} + 1$
$\therefore 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$

If $\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies \left(2 y\right) \left(0\right) = 2 x \implies x = 0 \implies y = 1$

So s there is just a single critical point when $x = 0$ so we need to look at the sign of $\frac{\mathrm{dy}}{\mathrm{dx}}$ for $x < 0$ and $x > 0$

First note that as $y = \sqrt{{x}^{2} + 1} \implies y \ge 1 \forall \in \mathbb{R}$, as ${x}^{2} \ge 0$

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{y}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{\sqrt{{x}^{2} + 1}}$

So we can see immediately that $\frac{\mathrm{dy}}{\mathrm{dx}}$ takes the sign of $x$, so we can conclude that:

If  { (x<0, => f'(x)<0, =>, "strictly decreasing"), (x=0, => f'(x)=0, =>, "stationary"), (x>0, => f'(x)>0, =>, "strictly increasing") :}

And as there is a single stationary point at $\left(0 , 1\right)$ this stationary point has to be a minimum.

Which concludes the deduction