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!