# Chaitanya's Random Pages

## August 28, 2015

### The discriminant trick

Filed under: mathematics — ckrao @ 12:22 pm

Suppose you wish to find the maximum value of $y =\frac{8}{2x+1} - \frac{1}{x}$ for $x > 0$. One way to do this without calculus is to massage the expression until one can apply an elementary inequality such as $u + \frac{1}{u} \geq 2$. To apply this particular result we aim to minimise $1/y$ and apply polynomial division.

\begin{aligned} \frac{1}{y} &= \frac{1}{\frac{8}{2x+1} - \frac{1}{x}}\\ &= \frac{x(2x+1)}{8x - (2x+1)}\\ &=\frac{2x^2+x}{6x-1}\\ &= \frac{(6x-1)(x/3 + 2/9) + 2/9}{6x-1}\\ &= \frac{1}{9}\left( 3x+2 + \frac{2}{6x-1} \right)\\ &= \frac{1}{9}\left( \frac{5}{2} + \frac{6x-1}{2} + \frac{2}{6x-1} \right)\\ &\geq \frac{1}{9}\left(\frac{5}{2} + 2 \right)\\ &= \frac{1}{9}\left(\frac{9}{2}\right)\\ &= \frac{1}{2}\quad\quad \quad \quad \quad (1) \end{aligned}

In the above steps we assume $6x-1 > 0$, otherwise $y$ is non-positive for $x > 0$. Hence $y \leq 2$ with equality when $\frac{6x-1}{2} = 1$ or $x = 1/2$.

Another elementary way that applies to quotients of quadratic polynomials is to re-write the expression as a quadratic in $x$:

\begin{aligned} y &= \frac{8}{2x+1} - \frac{1}{x}\\ x(2x+1)y &= 8x - (2x+1)\\ 2x^2y + (y-6)x + 1 &= 0.\quad \quad \quad \quad \quad \quad (2) \end{aligned}

For fixed $y$ this quadratic equation will have 0, 1 or 2 solutions in $x$ depending on whether its discriminant is negative, zero, or positive respectively. At any maximum or minimum value $y_0$ of the function, the discriminant will be zero since on one side of $y_0$ the quadratic equation will have a solution (discriminant non-negative) while on the other it will not (discriminant negative). In the image below a maximum is reached at $y_0 = 2$ while it is of opposite signs either side of this.

Hence setting the discriminant of the left side of (2) to 0, $(y-6)^2 - 4(2y) = 0$ from which $y^2 - 20y + 36 = (y-18)(y-2) = 0$. Hence extrema are at $y=2$ and $y = 18$. We can solve $(y-18)(y-2) \geq 0$ to find that $y \leq 2$ or $y \geq 18$ is the range of the function $y = \frac{8}{2x+1} - \frac{1}{x}$. This tells us that $y = 2$ is a local maximum (illustrated above) and $y = 18$ is a local minimum (occurring when $x < 0$).

One advantage of this method is that unlike elementary calculus, one bypasses the step of finding the corresponding $x$ value (i.e. by solving $dy/dx = 0$) before substituting this into the function to find the extremum value for $y$.

Another advantage is that the equation need not be polynomial in $y$. For example below is a plot of $x^2 + 4x\sin y + 1 = 0$. Using the above-mentioned discriminant trick we solve $16 \sin^2 y - 4 \geq 0$ and find the range of the function is when $\displaystyle \sin^2 y \geq 1/4$, or $y \in \bigcup_{k \in \mathbb{Z}} [k\pi + \pi/6, k\pi + 5\pi/6]$. Below is a plot confirming this using WolframAlpha.

The reader is encouraged to try out other examples, for example this method should work for any equation of the form $\displaystyle h(y) = \frac{ax^2 + bx + c}{dx^2 + ex + f}$ where $h$ is a continuous function of $y$. Of course one should also take care in noting when the function is defined before cross-multiplying.