HOW TO FIND INTERVALS IN A FUNCTION THAT INC. AND DEC.

[CHAMI]

I AM TAKE CAL AND IN ABOUT 3RD CHAPTER.

I KNOW HOW TO FIND DERIVITIVE OF THE FUNCTION AND SET IT EQUAL TO ZERO TO FIND THE CRITICAL NUMBERS. AFTER THAT I DO NOT HAVE A CLUE!

My fuction is this

f(x) = x^2 - 6x

I found the dervitive of this which is f '(x) = 2x -6
I set it = 0

2x - 6 = 0
x = 3//

Now that is the critical number and how do I find the intervals?
How do I find the relative maximum.

Every detail important for me and I am glad if you could explain like A, B, C...

Thank you. I am glad if you could show how to find intervals and relative minimum and max
THANKS

 

[stapel]

a) The critical numbers are the zeroes of the first derivative.

b) Given critical numbers a < b < c, etc, the intervals are then (-infinity, a), (a, b), (b, c), etc.

c) Pick any number within a given interval. Plug it into f'(x), and note the sign on the result.

d) If the sign changes from one interval to the next, then the critical point is an extrema. If the sign changes from negative to positive, then the extrema is a minimum; if the sign changes from positive to negative, then the extrema is a maximum.

Eliz.

 

[Gene]

You should know from algebra that this is an up-opening parabola. When you found that 2*3-6 = 0 you found that the minimum (vertex) is at x=3. Plug the 3 into the first equation to find that y = 3²-6*3 = -9 so the vertex is at (x,y)=(3,-9)
That is the only interesting point on a parabola since it goes up from there as x goes in either direction, plus or minus and x just keeps going too.

 

[CHAMI]

a) The critical numbers are the zeroes of the first derivative.

b) Given critical numbers a < b < c, etc, the intervals are then (-infinity, a), (a, b), (b, c), etc.

c) Pick any number within a given interval. Plug it into f'(x), and note the sign on the result.

d) If the sign changes from one interval to the next, then the critical point is an extrema. If the sign changes from negative to positive, then the extrema is a minimum; if the sign changes from positive to negative, then the extrema is a maximum.

Eliz.

What do you mean by sign changes? Do you mean when the sign change when you put okay critical number is 3 in my answer after I got the derivitive.

Okay how do you figure out interval number how do I know which number to put in the first derivitive I am confused now!

Totally with intervals. I guess I am glad if you can show.
:shock: I will try to figure out too, meanwhile I am glad if someone can show the process right after from derivitive to find intervals, relative min and max which is extrema i guess. Thanks


Thank you!

 

[CHAMI]

Okay when x= 3 in my question, how do I find out intervals? It a<b<c I do not know how to do that either! :shock: If you have x =2, x= 0 , x= 1 then you can find then how do you figure out increase and decrease?

 

[CHAMI]

:idea: I finally able to figure out to find increase and decrease of the function and relative min and max when you have two critical numbers. I draw the number line and get a test value and put that in to the first derivitive and look for the resultant sign whether it is + or - + increase - decrease. What still confuses is that when I get a critical number like x = 3 how to figure out the increase and decrease intervals? Will that (infinity, 3) and (3, 0) something like that, how to figure that out is not clear yet! :roll: :roll: :roll: :roll: :roll: :roll: :roll: :roll: :roll:

 

[Gene]

Just look at f'. When you solve it for = 0 each answer is a place where f changes direction and the intervals are (usually)
start of domain to lowest zero,
lowest zero to next zero
...
highest zero to end of domain.
f' = x-3
This f' has only one zero, x=3 and the domain is (-oo,oo) so there are two intervals one going down (f' is negative (-oo,3)) and one going up (f' is positive (3,oo))
f= x³-12x
f' = 3x^2-12
This f' has two zeros, x=-2 & x=+2 and the domain is (-oo,oo) so there are three intervals one going up f' is positive (-oo,-2), one going down (f' is negative (-2,2)) and one going up (f' is positive (2,oo))