How do you graph $f\left(x\right)=\frac{2}{{x}^{2}+1}$ using holes, vertical and horizontal asymptotes, x and y intercepts?

zakownikbj
2022-09-21
Answered

How do you graph $f\left(x\right)=\frac{2}{{x}^{2}+1}$ using holes, vertical and horizontal asymptotes, x and y intercepts?

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Trace Arias

Answered 2022-09-22
Author has **6** answers

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

Since $x}^{2}+1>0\forall x\in \mathbb{R$ there exists no holes in f(x)

Also, $\underset{\text{x-> +-oo}}{lim}f\left(x\right)=0$

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

For a maximum or minimum value; f'(x)=0

$\therefore \frac{-4x}{{({x}^{2}+1)}^{2}}=0\to x=0$

$f\left(0\right)=\frac{2}{0+1}=2$

Since $f\left(0\right)=2$ is a maximum of f(x)

The critical points of f(x) can be seen on the graph below:

graph{2/(x^2+1) [-5.55, 5.55, -2.772, 2.778]}

Since $x}^{2}+1>0\forall x\in \mathbb{R$ there exists no holes in f(x)

Also, $\underset{\text{x-> +-oo}}{lim}f\left(x\right)=0$

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

For a maximum or minimum value; f'(x)=0

$\therefore \frac{-4x}{{({x}^{2}+1)}^{2}}=0\to x=0$

$f\left(0\right)=\frac{2}{0+1}=2$

Since $f\left(0\right)=2$ is a maximum of f(x)

The critical points of f(x) can be seen on the graph below:

graph{2/(x^2+1) [-5.55, 5.55, -2.772, 2.778]}

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