Question

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

Answer 1

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.

Finally:

3000th ANSWER!!

Hopefully this helps!