Calculus
posted by dee .
relative extrema
x^42x^2+5
so far I know how to find the derivative which is
4x^32x now I am stuck...
Please help
Huh?
4x^32x=0
2x(2x^21)=0
and what are the roots? I will give you a hint: one is x=0 The other two are in the second parenthesis.
Is it x1 and x=1
I did a typo
Is it
x1, x+1
If you are in calc, you might want to consider some alternatives. This is pretty low level algebra. You are going to suffer with this lack of algebra understanding.
The roots of
2x(2x^21)=0
come from 2x=0 and 2x^21=0
or x=0 or x= + sqrt (1/2)
Three roots.
Good luck.
The way that I did the problem was different than the way you did it. I had 4x(x^21) = 0
That is how I came up with the roots of x=0, x=1, x=1
Was this way correct also?
No. You cannot factor 4x^32x the way you did. The only common factor is 2x.
4x^32x = 2x(2x^21)
The original problem was this
x^42x^2+5
Derivative = 4x^34x=0 when you factor you get 4x(x^21) = 0
Then you are left with x=0, x=1, x=1.
I think you are given me the wrong answer.

he was starting from where you left off with your derivative. the first time you wrote "so far I know how to find the derivative which is
4x^32x now I am stuck...
Please help "
that derivative is wrong. it is simply the one he used, hence the wrong answer. 
24X=260*(1.05^x/2)