critical numbers, extrema

[abby_07]

i need help with three different type of problems i started them except i get stuck. I can not finish the rest.

i need help in finding the critical numbers and the open intervals on which the function is increasing and decreasing.

f(x)=(x+1)^3

what i did first is took the derivative
f'(x)=3(x+1)^2
then i set it equal to zero to find the critical numbers
f'(x)=3(x+1)^2=0
x=0,-1 (these are the critical numbers)
this is where i get stuck. How do i figure where it is increasing and decreasing?


Then i need help on finding the absolute extrema of the function on the closed interval

f(x)= x(x^2+1)^(1/2) interval [0,2]

first i took the derivative and set it equal to zero
f'(x)=1/(x^2+1)^(3/2)=0
although i did not get and critical numbers , did i do something wrong?
well i figured if i did not get any critical numbers then i should use the endpoints
so i got that (0,0) was the minimum and (2, 2 square root of 5/5) was the maximum is this correct?

Then i have to find the realtive extrema of the function. is this the same as absolute extrema, if not then i do not no how to start this problem.

g(x)=3/2sin(pi*x/2-1) interval [0,4]

i know this is a lot, but i really dont no how to finish the problems of if i am approching it the right way

 

[camlax38]

i need help with three different type of problems i started them except i get stuck. I can not finish the rest.

i need help in finding the critical numbers and the open intervals on which the function is increasing and decreasing.

f(x)=(x+1)^3

what i did first is took the derivative
f'(x)=3(x+1)^2
then i set it equal to zero to find the critical numbers
f'(x)=3(x+1)^2=0
x=0,-1 (these are the critical numbers)
this is where i get stuck. How do i figure where it is increasing and decreasing?

So it all looks ok to me
What I always found useful was a number line or putting f(x) and f'(x) into a graphing calculator like a TI-83 and viewing the table (put the calc. into ask mode so you can enter your own numbers)

So if you do a number line something like this (ignore the periods I had to use them to keep the numbers separate

---------------------------------------------------
-1...............................0
Then we pick some numbers to plug in lets say -5, well when we pick x=-5 our function is -
then lets pick a number btwn -1 and 0, say -.5. Then our function will be +
Lastly lets pick a number larger than 0, say 5, we see our function is still positive.

So on our number line it would look like this

- ...............+......................+
---------------------------------------------------
...... -1................. 0

So we see that any number lower than -1 our function is neg, or decreasing greater than -1 our function is increasing

So you could say your function is decreasing on the interval (-infinity, -1) and is increasing on (-1, infinity)

hope that helps

 

[camlax38]

Then i need help on finding the absolute extrema of the function on the closed interval

f(x)= x(x^2+1)^(1/2) interval [0,2]

first i took the derivative and set it equal to zero
f'(x)=1/(x^2+1)^(3/2)=0
although i did not get and critical numbers , did i do something wrong?
well i figured if i did not get any critical numbers then i should use the endpoints
so i got that (0,0) was the minimum and (2, 2 square root of 5/5) was the maximum is this correct?

Then i have to find the realtive extrema of the function. is this the same as absolute extrema, if not then i do not no how to start this problem.

g(x)=3/2sin(pi*x/2-1) interval [0,4]

i know this is a lot, but i really dont no how to finish the problems of if i am approching it the right way

So for this I would check your derivative, to find f'(x) you need the multiplication rule as well as chain rule.

So f(x)*g(x)= f'(x)*g(x) + f(x)*g'(x)

In your case though g(x)= (x^2+1)^(1/2) you need to use chain rule for this
so ex> d/dx h(p(x))= h'(p(x))*p'(x)

As far as the extrema I think that would be right, but If you have to do it for a different type of problem
When f'(x)=0 and f"(x)>0 you have a min and when f"(x)<0 you have a max