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?
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
$$x = \frac{1 \pm \sqrt{0}}{2\frac{1}{2}} = 1.$$
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.
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 $\dfrac {1}{2}$.
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.
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 $\frac{1}{2}$.
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.
