Question

f(x)=(x^2-4)/(x^2+1) Find any asymptotes, extrema, intercepts, and points of inflection.

Answer 1

I've found an asymptote at x=1, x-intercepts at -2 and 2, and a y-intercept at -4. I'm having trouble finding the extrema...

Answer 2

\[f'(x)= (x^2+1)(2x)-(x^2-4)(2x)/(x^2+1)^2\]

Answer 3

which then equals \[2x^3+2x-2x^3+8x/(x^2+1)^2\]

Answer 4

\[\frac{x^2-4}{x^2+1}=1-\frac{5}{x^2+1} \]

Answer 5

What is that?

Answer 6

A plot is attached.

Answer 7

\[\lim_{x\to \infty } \, \left(1-\frac{5}{x^2+1}\right)=1 \]
Same limit as x approaches -Infinity

Answer 8

Set the 2nd derivative of f(x) to zero and solve for x. Plug the x's into f(x) to find the corressponding y's for the inflection points.\[\left\{\pm \frac{1}{\sqrt{3}},-\frac{11}{4}\right\} \]

Answer 9

Okay, thank you! But I am having trouble finding the extrema still...

Answer 10

It appears that I have a minimum at x=-4, but I need to find it by doing the math.

Answer 11

Set the first derivative to zero\[\frac{10 x}{\left(x^2+1\right)^2}=0 \]and solve for x.\[x=0 \] if x=0 then the corresponding y is -4 using\[\frac{x^2-4}{x^2+1} \]

Answer 12

THANK YOU! I don't know why I couldn't do that earlie!

Answer 13

You are very welcome.

Answer 14

Is \[-2x(x^2+2)/(x^2+1)^2\] the correct second derivative?

Answer 15

The second derivative is:\[-\frac{10 \left(3 x^2-1\right)}{\left(x^2+1\right)^3} \]