2
$\begingroup$

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?

$\endgroup$
14
  • $\begingroup$ 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
2
$\begingroup$

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 $x=6$.

$\endgroup$
2
  • $\begingroup$ 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
1
$\begingroup$

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

$$f(x)=x^{1/2}+2x^{-1/2},$$

$$f'(x)=\frac12x^{-1/2}-x^{-3/2},$$

$$f''(x)=-\frac14x^{-3/2}+\frac32x^{-5/2}.$$

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.