# Math Help - Question on monotonicity problem

1. ## Question on monotonicity problem

So here's my problem:

Find where the given function is increasing and decreasing.

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

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

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

So to find the split points would I solve $2x^2-2x>0$ and $2x^2-2x<0$ ?

2. Solve f'(x) = 0, this means $2x^2-2x=0$ which is a 2nd degree polynomial with a = 2 > 0, this means $2x^2-2x$ 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 $(x^2)^2=x^4$ is always positive so it doesn't interfere with the sign of your derivative

3. Can solved either with l'Hopital's rule.

4. My mistake! Sorry!

5. 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,

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

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.

6. Originally Posted by ascendancy523
So here's my problem:

Find where the given function is increasing and decreasing.

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

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

...
The derivative of a fraction is...

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

... so that is...

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

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

Kind regards

$\chi$ $\sigma$

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

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

... so that is...

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

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

Kind regards

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

Thank you everyone for the help!

8. You're very welcome. Have a good one!