Hopefully a simple f'(x) sign diagram question...

**calcconfused** · March 25th 2009, 07:52 AM

my teacher did a bad job of explaining how to do the sign diagrams for f'(x).

i know that you need to find the critical points and then put them on a number line. what i dont understand is how you figure out where its rising, falling, and where to do a sign change (all on the number line)

hopefully i worded that question right and i can get some help!

**Superjones** · March 25th 2009, 12:01 PM

it would be sine not sign lol. ummm you have to find the critical numbers that fit into the restricted domain of the sine funtion. example, you dont want to find all the critical numbers up to 12pie lol. you take the critical numbers you have... lets say your crit numbers are pie/2 and pie/6. you would have 2 crit numbers. so you would pick a number smaller than pie/6 like 0.... a number between pie/6 and pie/2 like pie/4 and then pick anumber bigger than pie/2 like pie. plug those three numbers in. and if you get a positive or neg number put a + or - in the space. + means the graph is going up / and - means the graph is going down \. this gives your the relative max's and mins. sorry i cant explain better. dont know computer sighns lol

**calcconfused** · March 25th 2009, 01:46 PM

Originally Posted by Superjones

it would be sine not sign lol. ummm you have to find the critical numbers that fit into the restricted domain of the sine funtion. example, you dont want to find all the critical numbers up to 12pie lol. you take the critical numbers you have... lets say your crit numbers are pie/2 and pie/6. you would have 2 crit numbers. so you would pick a number smaller than pie/6 like 0.... a number between pie/6 and pie/2 like pie/4 and then pick anumber bigger than pie/2 like pie. plug those three numbers in. and if you get a positive or neg number put a + or - in the space. + means the graph is going up / and - means the graph is going down \. this gives your the relative max's and mins. sorry i cant explain better. dont know computer sighns lol

im not sure you understood my question...

**Pinkk** · March 25th 2009, 01:56 PM

Critical points are where $f'(x)=0$ or where $f'(x)$ is undefined. Let's look at $f(x)=x^{2}$

$f'(x)=2x$ and $f'(x)=0$ when $x=0$ , so that is a critical point. So now let's take a value to the "left" of 0 (any negative number). Let's just pick $x=-1$ for simplicity. $f'(-1)=-2$ , which means that when $x<0$ , $x^{2}$ is decrease (you can look at the graph to verify[/tex]. If $f'(x)<0$ , that means $f(x)$ is decreasing. If we pick 1, we'll see the opposite.

Remember: If $f'(x)$ is negative on some interval, then $f(x)$ is decreasing on that interval, and the opposite is true. And if $f'(x)=0$ , or if $f'(x)$ is undefined at some point, then that point is a critical point, and could be a local extreme.

A simple sign diagram for $f'(x)=2x$

A diagram that may be helpful:

**calcconfused** · March 25th 2009, 02:29 PM

Originally Posted by Pinkk

Critical points are where $f'(x)=0$ or where $f'(x)$ is undefined. Let's look at $f(x)=x^{2}$

$f'(x)=2x$ and $f'(x)=0$ when $x=0$ , so that is a critical point. So now let's take a value to the "left" of 0 (any negative number). Let's just pick $x=-1$ for simplicity. $f'(-1)=-2$ , which means that when $x<0$ , $x^{2}$ is decrease (you can look at the graph to verify[/tex]. If $f'(x)<0$ , that means $f(x)$ is decreasing. If we pick 1, we'll see the opposite.

Remember: If $f'(x)$ is negative on some interval, then $f(x)$ is decreasing on that interval, and the opposite is true. And if $f'(x)=0$ , or if $f'(x)$ is undefined at some point, then that point is a critical point, and could be a local extreme.

A simple sign diagram for $f'(x)=2x$

A diagram that may be helpful:

that helped a pretty good deal...wouldnt it be 0 and 2 on the number line though? i also dont understand when to change between positive and negative. my professor said to start out with the right most interval and move left. he said to put an x on the number if the factor has an odd power and THIS is where you change from + to - or vice versa.

**Pinkk** · March 25th 2009, 02:35 PM

$x=2$ is not a critical point.

To find change of signs, you test values to the left of the critical point and values right of the critical point and check their signs. I'm not sure what you're professor is telling you. Either you're not explaining it clear or your professor is using a method I have never seen before.

**Pinkk** · March 25th 2009, 02:54 PM

Let's look at a function where there are two critical points. $f(x)=\frac{x^{3}}{3}-\frac{x^{2}}{2}$

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

$f'(x)=0$ when $x=0,x=1$ . So you have two critical points.

So let's set up our sign diagram:

So we will need to test three values; one that is less than 0, one that is between 0 and 1, and one greater than 1. They can be any values so long as they fit those requirements.

$f'(-1)=2$
$f'(\frac{1}{2})=-\frac{1}{4}$
$f'(2)=2$

So now we write in the following into our sign diagram: