# Question on monotonicity problem

• Jun 26th 2010, 11:24 AM
ascendancy523
Question on monotonicity problem
So here's my problem:

Find where the given function is increasing and decreasing.

$\displaystyle f(x)=\frac{x-1}{x^2}$

So $\displaystyle f'(x)=\frac{(x-1)(2x)-(x^2)(1)}{(x^2)^2)}$

$\displaystyle =\frac{2x^2-2x-x^2}{(x^2)^2)}$
$\displaystyle =\frac{x^2-2x}{(x^2)^2}$

So to find the split points would I solve $\displaystyle 2x^2-2x>0$ and $\displaystyle 2x^2-2x<0$ ?
• Jun 26th 2010, 11:45 AM
Utherr
Solve f'(x) = 0, this means $\displaystyle 2x^2-2x=0$ which is a 2nd degree polynomial with a = 2 > 0, this means $\displaystyle 2x^2-2x$ is negative between its roots [x1, x2] and positive for the rest (-inf, x1) U (x2, +inf), if a < 0 the sign would be reversed.

if f(x) < 0 then f(x) is decreasing on that interval, f(x) > 0 increasing on that interval

i forgot to mention $\displaystyle (x^2)^2=x^4$ is always positive so it doesn't interfere with the sign of your derivative
• Jun 26th 2010, 11:52 AM
Also sprach Zarathustra
Can solved either with l'Hopital's rule.
• Jun 26th 2010, 11:55 AM
Also sprach Zarathustra
My mistake! Sorry!
• Jun 26th 2010, 12:40 PM
Ackbeet
I think you have a sign error in your derivative. For the quotient rule, I always use this mnemonic:

"lo dee-hi minus hi dee-lo over the square of what's below". That is,

$\displaystyle \left(\frac{f(x)}{g(x)}\right)'=\frac{g(x)f'(x)-f(x)g'(x)}{g^{2}(x)}$.

If you compare this with how you took your derivative, I think you'll see the error. And this error will affect your results, because all your answers will be reversed from what they should be.
• Jun 27th 2010, 07:12 AM
chisigma
Quote:

Originally Posted by ascendancy523
So here's my problem:

Find where the given function is increasing and decreasing.

$\displaystyle f(x)=\frac{x-1}{x^2}$

So $\displaystyle f'(x)=\frac{(x-1)(2x)-(x^2)(1)}{(x^2)^2)}$

...

The derivative of a fraction is...

$\displaystyle \displaystyle \frac{d}{dx} \frac{f(x)}{g(x)} = \frac{g(x)\ f'(x) - f(x)\ g'(x)}{g^{2}(x)}$ (1)

... so that is...

$\displaystyle \displaystyle \frac{d}{dx} \frac{x-1}{x^{2}} = \frac{2-x}{x^{3}}$ (2)

... and the function $\displaystyle \displaystyle \frac{x-1}{x^{2}}$ is increasing for $\displaystyle 0 < x < 2$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jun 27th 2010, 03:13 PM
ascendancy523
Quote:

Originally Posted by chisigma
The derivative of a fraction is...

$\displaystyle \displaystyle \frac{d}{dx} \frac{f(x)}{g(x)} = \frac{g(x)\ f'(x) - f(x)\ g'(x)}{g^{2}(x)}$ (1)

... so that is...

$\displaystyle \displaystyle \frac{d}{dx} \frac{x-1}{x^{2}} = \frac{2-x}{x^{3}}$ (2)

... and the function $\displaystyle \displaystyle \frac{x-1}{x^{2}}$ is increasing for $\displaystyle 0 < x < 2$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Alright, so f(x) is increasing on [0,2] and decreasing on ($\displaystyle -\infty$,0]$\displaystyle \cup$[2,$\displaystyle \infty$)?

Thank you everyone for the help!
• Jun 28th 2010, 05:21 PM
Ackbeet
You're very welcome. Have a good one!