Max Macias

2022-08-09

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

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 ${D}_{f(g(x))}=(-1,+\mathrm{\infty})$

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 $(-1;+\mathrm{\infty})$

We have g is increasing

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

So $g([-1;+\mathrm{\infty}))=[0,+\mathrm{\infty})$.

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.

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 ${D}_{f(g(x))}=(-1,+\mathrm{\infty})$

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 $(-1;+\mathrm{\infty})$

We have g is increasing

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

So $g([-1;+\mathrm{\infty}))=[0,+\mathrm{\infty})$.

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.

Kobe Ortiz

Beginner2022-08-10Added 9 answers

Explanation:

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 $[0,\mathrm{\infty})$. Therefore the variance is $(}h(0)-h(1){\textstyle )}+{\textstyle (}\underset{x\to \mathrm{\infty}}{lim}h(x)-h(1){\textstyle )}=\mathrm{\infty}.$

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 $[0,\mathrm{\infty})$. Therefore the variance is $(}h(0)-h(1){\textstyle )}+{\textstyle (}\underset{x\to \mathrm{\infty}}{lim}h(x)-h(1){\textstyle )}=\mathrm{\infty}.$

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