0
$\begingroup$

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!

  • 1
    No problem, I was speculating you meant this. However this is not$a$quadratic (grade 2) function at all, so it would definitely be inappropriate to apply the method. It's good though that you sought to use a simple method first! Some students just blindly use the sledgehammer...2012-05-26

2 Answers 2

1

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!

  • 0
    Thank you for your help! Makes sense :).. The fractions were confusing me, that's why I didn't try the extreme points from the start.2012-05-26
0

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.