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

1 Answer
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 + 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.

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Finally:

enter image source here
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Hopefully this helps!