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:


$$\frac{2x^{\frac{3}{2}} -3x^{}\frac{3}{2}+6x^{\frac{1}{2}}}{4x^{3}}=0$$ $$-x^{\frac{3}{2}} + 6x^{\frac{1}{2}} = 0$$ $$(-x^{\frac{3}{2}})^{2} + (6x^{\frac{1}{2}})^{2}$$ $$x^{3} + 36x = 0$$ $$x (x^{2} + 36) = 0$$ $$x = 0 \text{ or } x^{2} = - 36 \text{ no solution..}$$

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?

  • $\begingroup$ Never mind that. $\endgroup$
    – Cro-Magnon
    Apr 10 '16 at 21:00
  • $\begingroup$ Why do you say that this "doesn't seem right"? $\endgroup$ Apr 10 '16 at 21:01
  • $\begingroup$ Why should you expect to be able to solve it? Not every function has an inflection point. Take for example $\frac{1}{2}x^2$ whose second derivative is identically equal to $1$ and is never zero. $\endgroup$
    – JMoravitz
    Apr 10 '16 at 21:01
  • 1
    $\begingroup$ Perhaps you could show your work in the quotient rule, your numerator looks suspicious to me. It looks almost like you forgot to multiply by the $-2$. $\endgroup$ Apr 10 '16 at 21:02
  • 1
    $\begingroup$ Please include your work so that we can find where the error is. If you don't include your work, it's hard for us to help you without doing the problem. $\endgroup$ Apr 10 '16 at 21:06

We proceed via the quotient rule:


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 $x=6$.

  • $\begingroup$ Hey Thanks! I got the exact same thing as you did. But what If I do not factor out the $x^{1/2}$ and instead cancel the numerator (the lower part of the fraction) and then do the following: $(-x^{\frac{3}{2}})^{2} + (6x^{\frac{1}{2}})^{2}$ Is this valid? $\endgroup$
    – Cro-Magnon
    Apr 10 '16 at 21:16
  • $\begingroup$ In general, $a^2+b^2 \neq (a+b)^2$ $\endgroup$
    – zz20s
    Apr 11 '16 at 1:32

Not the answer that you expect, but a simple way to get the correct answer.




Multiplying by $-4x^{-5/2}$, the inflection point is at $x-6=0$.


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