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

Feb 12, 2018

Intercepts

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

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

$0 = 1 + \frac{1}{x} ^ 2$

$- 1 = \frac{1}{x} ^ 2$

$- {x}^{2} = 1$

${x}^{2} = - 1$

$x = \emptyset$

So no x-intercept either.

Finding local maximum/minimum

We differentiate:

$f ' \left(x\right) = - 2 {x}^{- 3}$

Now let's check for critical points.

$0 = - 2 {x}^{- 3}$

$x = 0$

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

Inflection points

$f ' ' \left(x\right) = 6 {x}^{- 4}$

$0 = 6 {x}^{- 4}$

$x = 0$

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

Asymptotes

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

End behavior

Now we must examine what $f \left(x\right)$ does at $0$, and $\pm \infty$.

${\lim}_{x \to \infty} 1 + \frac{1}{x} ^ 2 = 1$
${\lim}_{x \to - \infty} 1 + \frac{1}{x} ^ 2 = 1$
${\lim}_{x \to {0}^{+}} 1 + \frac{1}{x} ^ 2 = + \infty$
${\lim}_{x \to {0}^{-}} 1 + \frac{1}{x} ^ 2 = + \infty$

Graphing

We now conclude the graph must resemble the following.

Finally: