How do you sketch the curve $f(x)=1+1/x^2$ by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

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Answer 1 · 2018-02-12T16:09:00

Intercepts

There will be no $y$ intercept, because the function is rendered undefined when $x = 0$ .

There will be a x-intercept when $y = 0$ .

$0 = 1 + 1/x^2$

$-1 = 1/x^2$

$-x^2 = 1$

$x^2 = -1$

$x = O/$

So no x-intercept either.

Finding local maximum/minimum

We differentiate:

$f'(x) = -2x^(-3)$

Now let's check for critical points.

$0 = -2x^(-3)$

$x = 0$

But since $x = 0$ is undefined in the original function, we can't use this critical point.

Inflection points

$f''(x) = 6x^(-4)$

$0 = 6x^(-4)$

$x= 0$

Once again, as $x =0$ is undefined in the original function, we can't use it.

Asymptotes

We know from above that $x !=0$ and $y != 0$ , therefore these will be our asymptotes.

End behavior

Now we must examine what $f(x)$ does at $0$ , and $+- oo$ .

$lim_(x->oo) 1 + 1/x^2 = 1$
$lim_(x->-oo) 1 + 1/x^2 = 1$
$lim_(x-> 0^+) 1 + 1/x^2 = +oo$
$lim_(x-> 0^-) 1 + 1/x^2 = +oo$

Graphing

We now conclude the graph must resemble the following.

enter image source here

Finally:

enter image source here
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Hopefully this helps!

How do you sketch the curve f(x)=1+1/x^2f(x)=1+1/x^2 by finding local maximum, minimum, inflection points, asymptotes, and intercepts?