How can one determinate the variationt of $f(g(x))$

Question

Given the two functions :

$$f(x)=x^2-2x$$

$$g(x)=\sqrt{x+1}$$

The question is determinate the variation of $f(g(x))$

We have $D_{f(g(x))}=(-1,+\infty)$

For $f$ it's decreasing for $x<1$

Increasing for $x>1$

For $g$ its increasing for $x>-1$

To determinate the variationf of $f(g(x))$

In interval $(-1;+\infty)$

We have $g$ is increasing

$X>-1$ means that $g(x)>g(-1)$ so $g(x)>0$

So $g([-1;+\infty))=[0,+\infty)$

But the problem is that $f$ in that interval is increasing and decreasing Im stuck here.

Can one write a methode to answer any question like this theoricaly .

Answer 1

You ought to find intervals on which $g(x)<1$ and on which $g(x)>1$ and then compute the variation on each interval separately. We have $g(1)=1$ and $g$ is increasing, so $h:=f\circ g$ is decreasing on $(-1,0]$ and increasing on $[0,\infty)$. Therefore the variance is $$\bigl(h(0)-h(1)\bigr) + \bigl(\lim_{x\to\infty} h(x)-h(1)\bigr) = \infty.$$