We have
$f:R\rightarrow R$
$f(x)=\frac{3x-1}{3x^2+1}$
Determine $x$ for which the function has the maximum value. How can I determine the maximum of this function?
Thank you very much in advance!
We have
$f:R\rightarrow R$
$f(x)=\frac{3x-1}{3x^2+1}$
Determine $x$ for which the function has the maximum value. How can I determine the maximum of this function?
Thank you very much in advance!
You can first observe that it has a horizontal asymptote at y=0, so it is not going to go up or down forever on the ends.
In this case, it's probably best to go straight for the derivative and find the critical points $x=1$ and $x=-1/3$. Doing the first or second derivative test tells you the max occurs at $x=1$, and so that maximum value of $f$ is $1/2$.
Fill in the gaps!
I don't know what you mean by "$a>0$", but you can solve this by just taking the derivative and setting it equal to zero:
$f'(x) = \frac{(3x^2 + 1) 3 - (3x-1)(6x)}{(3x^2 +1)^2} = \frac{-9x^2 + 6x +3}{(3x^2+1)^2}$
$= -3 \frac{3x^2 - 2x - 1}{(3x^2 + 1)^2}$
So just find the roots of $3x^2 - 2x - 1=0$, assuming my algebra doesn't have any mistakes. Take whichever one has a larger value when you plug into $f$, and then show that it is, in fact, a maximum.