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

##### 1 Answer

**Intercepts**

There will be no

There will be a x-intercept when

#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

**Inflection points**

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

#0 = 6x^(-4)#

#x= 0#

Once again, as

**Asymptotes**

We know from above that

**End behavior**

Now we must examine what

#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!