# First and second derivatives

1. Nov 1, 2008

### Swerting

This problem is fairly straightforward. I have a function and domain:
$$f(x)=2ln(x^{2}+3)-x -3\leqx\leq5$$
The question eventually asks for the inflection points, which I know occur when f''(x) is zero or undefined.
I found the first derivative and second derivative to be, respectively:
$$f'(x)=(4x)(x^{2}+3)^{-1}-1$$
$$f"(x)=(4)(x^{2}+3)^{-1}-(4x)(x^{2}+3)^{-2}$$
I have double checked my math and can't seem to find anything wrong with it.
Also, I have looked at a calculator and it shows that the original function does have two inflection points, but when I try to set f"(x) to zero or to be undefined (devide by zero), I always end up with an impossible solution. I don't beleive that this involves irrational numbers, so I appreciate any help at all.

2. Nov 1, 2008

### HallsofIvy

Staff Emeritus

You have
$$f"(x)=(4)(x^2+3)^{-1}-(4x)(x^2+3)^{-2}$$

With $f'(x)= (4x)(x^2+3)^{-1}- 1$ the second derivative is
$$4(x^2+ 3)^{-1}- (4x)(x^2+ 3)^{-2}(2x)= 4(x^2+ 3)^{-1}- 8x^2(x^2+3)^{-2}$$
$$= \frac{4(x^2+ 3)}{(x^2+ 3)^{-2}}- \frac{8x^2}{(x^2+ 3)^{-2}}= \frac{-8x^2+ 3}{(x^2+ 3)^{-2}}$$
The second derivative always exists but there are two places where it is 0.

3. Nov 2, 2008

### Swerting

Thank you very much, I see it now.
I figured it was something in the calculation of the first to second, I just couldn't see it.