<s>[INST] Why can't I solve for this second derivative? Here's the equation I have to find the second derivative point for.
$$f(x)=\frac{x+2}{x^{\frac{1}{2}}}$$
$$f'(x) = \frac{x-2}{2x^{\frac{3}{2}}}$$
From here I then calculate the second derivative and set it equal to 0.
But it doesn't work.. Take a look:
$$f''(x)=\frac{-x^\frac{3}{2} + 6x^{\frac{1}{2}}}{4x^{3}} = 0$$
FROM first DERIVATIVE TO second:
$$\frac{2x^{\frac{3}{2}}[x-2]'-((x-2)[2x^{\frac{3}{2}}]'}{4x^{3}}$$
$$\frac{2x^{\frac{3}{2}} -3x^{}\frac{3}{2}+6x^{\frac{1}{2}}}{4x^{3}}=0$$
$$-x^{\frac{3}{2}} + 6x^{\frac{1}{2}} = 0$$
$$(-x^{\frac{3}{2}})^{2} + (6x^{\frac{1}{2}})^{2}$$
$$x^{3} + 36x = 0$$
$$x (x^{2} + 36) = 0$$
$$x = 0  \text{   or    } x^{2} = - 36 \text{      no solution..}$$
This doesn't seem right.. Yet I have no idea why. It's easier to write the function as a product but I want to solve it using the quotient rule..
What's going on?
 [/INST] We proceed via the quotient rule:
$$f''(x)=\frac{2x^{3/2}-(x-2)3x^{1/2}}{4x^3}=\frac{2x^{3/2}-3x^{3/2}+6x^{1/2}}{4x^3}=x^{1/2}\left(\frac{6-x}{4x^3}\right)=\frac{6-x}{4x^{5/2}}$$
We can set $f''(x)=0$ and find that $f''(6)=0$. To find the points of inflection, we simply need to test a point less than and greater than $x=6$.
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