determining the maximum of a rational function I was going over some practice problems and got stuck with this one: I am supposed to find the maximum of the function: 

$$\dfrac{x}{x^2+1}$$ 

on the interval $(0,4)$.
All I can think of is to try and plug in values from the given interval . Is there a more systematic approach to finding maxima of rational functions?
 A: There is a way to do this without calculus.
You want to find the minimum value of $a$ for which $f(x) \leq a$ for all $a$. In other words, the minimum value of $a$ for which
$$\frac{x}{x^2+1} \leq a$$
for all $x\in [0,4]$. Since $x^2+1$ is positive, the above inequality is equivalent to
$$x\leq ax^2+a$$
$$ax^2-x+a \geq 0.$$
Call $g(x,a) = ax^2-x+a$. If $g(x,a)$ is always positive for $x$ in $[0,4]$, this means that $g(x,a)$ never crosses the $x$ axis: it has either no real roots in $g(x,a)$, or a double root. In both these cases, you then need to check if $g(x,a)$ is always nonnegative (instead of always nonpositive). (You probably covered how to solve polynomial inequalities in precalculus).
The roots of $g(x,a)$ are
$$x = \frac{1 \pm \sqrt{1-4a^2}}{2a}.$$
If $a\geq\frac{1}{2}$, then the roots are double or not real, and $g(0,a) = a \geq 0$, so all $a$ in this region are solutions to the inequality.
If $a\leq \frac{-1}{2}$, then the roots also are double or not real, but this time $g(0,a) < 0$ and the inequality is always violated.
If $-\frac{1}{2} < a < \frac{1}{2}$, then there is a non-doubled real root in $[0,4]$ and so the inequality does not hold for all $x$ in $[0,4]$.
So, finally, we've shown that $f(x) \leq a$ for all $x$ in $[0,4]$ precisely when $a \geq \frac{1}{2}$, so the maximum of $f$ in that region must be $\frac{1}{2}$. To find the value of $x$ that gives the maximum, we plug $a=\frac{1}{2}$ into the expression for the roots above and get
$$x = \frac{1 \pm \sqrt{0}}{2\frac{1}{2}} = 1.$$
A: Here's one way:
$$
\frac{x}{x^2+1} = \frac{1}{\left(\frac{x^2+1}{x}\right)}
$$
so the problem is that of minimizing
$$
\frac{x^2+1}{x} = x + \frac 1 x = \left(x - 2 + \frac 1 x \right) + 2 = \left(\sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 + 2.
$$
That square is minimized by making to equal to zero.
It can also be done by using calculus.  And I think there may be a proof via Euclidean geometry with essentially no algebra, but I'm not sure of that yet.
A: Try to find the derivative of the function and to determine when the function is decreasing or increasing.
This will give you all the information you want.
Hint :the value is $$f(1)=\frac{1}{2}$$  
EDIT:
Find the minimum of the inverse:$\dfrac{x^2+1}{x}=x+\dfrac{1}{x}$
But If $x>0$, then $x+\dfrac{1}{x}\geq2$ which means that the minimum value of the quantity you want is $2$ or the maximum of the quantity is $\dfrac {1}{2}$.
A: Hint:

*

*Find the critical points in your interval(values of $x$ at which $f'(x)=0$)

*Evaluate $f$ at the critical points in 1 and at $x=0,4$

*The maximum value of $f$ in 2 is the required value.

A: How about trigonometry?  Set up a right triangle with a "vertical leg" of length 1 and a "horizontal leg" of length $ \ x \ $ ; the hypotenuse then has length $ \ \sqrt{x^2 + 1 } \ . $  With the angle "opposite" the vertical leg being $ \ \theta \ $ , we have
$$ \frac{x}{x^2 + 1} \ = \ \frac{x}{\sqrt{x^2 + 1}} \ \cdot \ \frac{1}{\sqrt{x^2 + 1}} \ \rightarrow \ \sin \theta \cdot \cos \theta \ = \ \frac{1}{2} \sin 2 \theta \ \ , $$
which has a maximum value of $ \ \frac{1}{2} \ . $
This occurs when $ \ 2 \theta \ = \ \frac{\pi}{2} \ \Rightarrow \ \theta \ = \ \frac{\pi}{4} \ . $  This gives us a right isosceles triangle, which requires that $ \ x = 1 \ . $
As for a general method of finding extrema of rational functions of polynomials... well, that's what differential calculus is for.  (The approach I've taken will only work for certain rational functions...)
EDIT -- I now spot  lab bhattacharjee's link in a comment above to a past problem which uses a closely related approach on a very similar function.
A: Not sure if this is a general solution that will always work, but this problem is in Precalculus so it was chosen to be able to be done using rational function concepts. It is a simplified (less rigorous) version of an answer already posted  by user7530.
If there is a y-value maximum of $f(x)$ then there should be an x-value maximum of the inverse. So there should be a restriction on the domain of $f^{-1}(x)$ to make a maximum. So lets find the inverse of $f(x)$ and then see if there are any domain restrictions on it.
$$
\eqalignno{
x&=\frac{y}{y^2+1}  &\text{get $y^2+1$ out of denominator}  \\  
xy^2+x&=y   &\text{get set to 0 so we can solve for y} \\  
xy^2-y+x&=0  
}$$
We still need to solve for y so we can find the inverse. We can use the quadratic formula, where a = x, b = -1, and c = x.
So
$$
f^{-1}(x)=y=\frac{1\pm\sqrt{1-4x^2}}{2x}
$$
Which means that $1-4x^2\ge0$ would be a restriction on the domain of the inverse. When we solve for x, $x \le \frac{1}{2}$. So since the maximum x value in the inverse is $\frac{1}{2}$, the maximum y value for the function $f(x)$ must be $\frac{1}{2}$.
A: assuming this qn doesn't allow calculus,
$(x-1)^2\geq 0$
$x^2+1\geq 2x$
$\frac{x}{x^2+1}\leq\frac{1}{2}$
max is achieved at $x=1$
however, "the more systematic approach part" probably requires calculus.
