# Math Help - graphing rational function

1. ## graphing rational function

just wondering if anyone knows how to determine the inflection points,max/min and concavity of the rational function:

f(x)=x2+4/x2-9

just finished finding the vertical and horizontal asymptotes, symmetry, x and y intercepts, and domain, but can't recall the meathods for determining the above. I used to know how to do this like the back of my hand!

2. Originally Posted by dcfi6052
just wondering if anyone knows how to determine the inflection points,max/min and concavity of the rational function:

f(x)=x2+4/x2-9

just finished finding the vertical and horizontal asymptotes, symmetry, x and y intercepts, and domain, but can't recall the meathods for determining the above. I used to know how to do this like the back of my hand!
You do it just like you do it for any other function.

Relative max/min:
Set f'(x) = 0. The values of x where this happens are your candidates.

$f'(x) = \frac{(2x)(x^2-9) - (x^2+4)(2x)}{(x^2-9)^2}$

$f'(x) = -\frac{26x}{(x^2-9)^2}$

So f'(x) = 0 only when x = 0.

Now is this a relative max or a relative min? You can find out by graphing, but we need the second derivative to get the other stuff anyway, so let's do the second derivative test on this.

$f''(x) = \frac{(-26)(x^2-9)^2 - (-26x)(2)(x^2-9)(2x)}{(x^2-9)^4}$

$f''(x) = \frac{78(x^2+3)}{(x^2-9)^3}$ (after a bit of factoring)

So what is f''(x) for x = 0? $f''(0) = -\frac{26}{81}$. Since f''(0) is negative, the point (0, -4/9) is a relative maximum.

As it happens, the second derivative IS the concavity, so you have already found this.

An inflection point is where f''(x) = 0. So this happens when $x^2 + 3 = 0$, which has no real solutions for x. Thus there are no inflection points.

-Dan