# How can one determinate the variationt of f(g(x))? Given the two function: f(x)=x^2-2x, g(x)=sqrt{x+1}

How can one determinate the variationt of f(g(x))
Given the two functions:
$f\left(x\right)={x}^{2}-2x$
$g\left(x\right)=\sqrt{x+1}$
The question is determinate the variation of f(g(x))
We have ${D}_{f\left(g\left(x\right)\right)}=\left(-1,+\mathrm{\infty }\right)$
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 $\left(-1;+\mathrm{\infty }\right)$
We have g is increasing
$X>-1$ means that $g\left(x\right)>g\left(-1\right)$ so $g\left(x\right)>0$.
So $g\left(\left[-1;+\mathrm{\infty }\right)\right)=\left[0,+\mathrm{\infty }\right)$.
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.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

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