# Stationary points, local maxima, minima of a function

• Aug 2nd 2010, 12:12 PM
cozza
Stationary points, local maxima, minima of a function
Hi,

Does anyone know how you work out the intervals on which a function is increasing or decreasing and any stationary points, local maxima and local minima?

The function I have is $f(x)=\frac{13-6x}{4-x^2}$

I have differentiated it and have $f '(x)=-\frac{6x^2-26x+24}{(4-x^2)^2}$

Which can be factorised to $\frac{2(-3x-4)(x-3)}{(2+x)(2-x)}$

From other eg's I think a sign table is the best way of doing this, but I'm not sure what the headings / columns should be.

From trial and error I know there is a stationary point at 3, i.e. f '(x) is 0 when x=3, but I'm not sure how to work out the other things.

I would be really grateful for any advice
• Aug 2nd 2010, 12:31 PM
yeKciM
your function is defined on $(-\infty, -2) \cup (2 , + \infty)$
that u'll know from $4-x^2 \neq0$

your zero point is $13-6x=0 \Rightarrow x=\frac {13}{6}$

there are like 7 thing's for u to do ti'll you are able to draw that function ....
u need first and second derivation .... and so on :D
are u familiar how (and why ) each of it is done ?

Edit :: lost while translate :D sorry :D (hello Ackbeet and thanks :D )

• Aug 2nd 2010, 12:34 PM
Ackbeet
I disagree. The function is defined on $\mathbb{R}\setminus\{-2,2\}.$ Stationary points are where the derivative is equal to zero. In setting a fraction equal to zero, you need only make sure that at those points, the numerator is zero, and the denominator is not zero. Use the second derivative test to find maxima and minima.
• Aug 2nd 2010, 12:43 PM
cozza
Quote:

Originally Posted by yeKciM
your function is defined on $(-\infty, -2) \cup (2 , + \infty)$
that u'll know from $4-x^2 \neq0$

your stationary point is $13-6x=0 \Rightarrow x=\frac {13}{6}$

there are like 7 thing's for u to do ti'll you are able to draw that function ....
u need first and second derivation .... and so on :D
are u familiar how (and why ) each of it is done ?

Hey. Thanks for the reply (Happy)

I have a strategy in my book I have been working from:
1. the domain of f (all real numbers except +/- 2)
2. whether f is odd or even (f is neither)
3. the x and y intercepts of f (x=13/6, y=13/4)
4. the intervals on which x is positive or negative (positive on (-2,0), 0, (0,2) and $(2, \infty)$ and negative on $(-\infty, -2)$
5. the intervals on which f is increasing / decreasing and any stationary points, local maxima, local minima - this is where I got stuck
6. the asymptotic behaviour of f

Is this what you meant? I haven't used the second derivative though. Do I need to?

Thanks again (Happy)
• Aug 2nd 2010, 12:54 PM
cozza
Quote:

Originally Posted by Ackbeet
I disagree. The function is defined on $\mathbb{R}\setminus\{-2,2\}.$ Stationary points are where the derivative is equal to zero. In setting a fraction equal to zero, you need only make sure that at those points, the numerator is zero, and the denominator is not zero. Use the second derivative test to find maxima and minima.

On differentiating

$f '(x)=-\frac{6x^2-26x+24}{(4-x^2)^2}$ again, I get

$f ''(x)=-\frac{6x^3-10x^2+24x-16}{(4-x^2)^4}$.

Does this look like I've differentiated in correctly? (Worried)
• Aug 2nd 2010, 12:57 PM
yeKciM
u need first derivation to find stationary points (sorry for that early) , then u find second derivation and when u equal it with zero u'll get saddle point's (sorry i didn't find better word to translate that ) if there are any :D
but when u put your stationari point in second in second derivation u'll see if it is min or max :D when u put stationary point in second derivation and it's $f_{(x)} '' <0 \Rightarrow max$ ,or if it is $f_{(x)} '' >0 \Rightarrow min$ ...
• Aug 2nd 2010, 01:14 PM
yeKciM
Quote:

Originally Posted by cozza
On differentiating

$f '(x)=-\frac{6x^2-26x+24}{(4-x^2)^2}$ again, I get

$f ''(x)=-\frac{6x^3-10x^2+24x-16}{(4-x^2)^4}$.

Does this look like I've differentiated in correctly? (Worried)

$\displaystyle f'(x)= \frac {-6x^2-26x-24}{(4-x^2)^2} \Rightarrow 3x^2-13x-12= -(3x+4)(x+3)=0$ and you have your stationary points :D
• Aug 2nd 2010, 02:08 PM
cozza
Quote:

Originally Posted by yeKciM
$f'(x)= \frac {-6x^2-26x-24}{(4-x^2)^2} \Rightarrow 3x^2-13x-12= -(3x+4)(x+3)=0$ and you have your stationary points :D

so f'(x) is 0 when x=3 or x=-4/3

So if I now put those values of x into the second derivative if they are <0 then it is a local maxima and if >0 then it is a local minima?

I got that x=3 is a local maxima and x=-4/3 a local minima

For the asymptotic behaviour, since the denominator is 0 when x is +/- 2, I think the lines x=2 and x=-2 are vertical asymptotes. For horizontal asymptotes my book says "To detect horizontal asymptotes of a rational function, we divide bothe the numerator and denominator by the dominant term of the denominator and consider the behaviour as $x\rightarrow\infty$ and as $x\rightarrow-\infty$.

So, $f(x)=\frac{13-6x}{4-x^2}=\frac{\frac{13}{x^2}-\frac{6x}{x^2}}{\frac{4}{x^2}-1}$
$\rightarrow\frac{0-0}{0-1}=0 as x\rightarrow+/-\infty$, which means f has the horizontal asymptote y=0, but that doesn't work, according to what I think the graph should look like? (Headbang)
• Aug 2nd 2010, 02:22 PM
yeKciM
hehehehe... i got lost there working on second so i can't confirm to u is it or is it not correct that one u have done :D sorry
but there u are how your graphic should look like
• Aug 2nd 2010, 02:30 PM
cozza
That's what I get too. Thank you for all your help :)