What is the range of y=[(1-x)^(1/2)]/(2x^2+3x+1)?

May 22, 2015

First let us consider the domain:

For what values of $x$ is the function defined?

The numerator ${\left(1 - x\right)}^{\frac{1}{2}}$ is only defined when $\left(1 - x\right) \ge 0$. Adding $x$ to both sides of this you find $x \le 1$.

We also require the denominator to be non-zero.
$2 {x}^{2} + 3 x + 1 = \left(2 x + 1\right) \left(x + 1\right)$ is zero when $x = - \frac{1}{2}$ and when $x = - 1$.

So the domain of the function is

$\left\{x \in \mathbb{R} : x \le 1 \mathmr{and} x \ne - 1 \mathmr{and} x \ne - \frac{1}{2}\right\}$

Define $f \left(x\right) = {\left(1 - x\right)}^{\frac{1}{2}} / \left(2 {x}^{2} + 3 x + 1\right)$ on this domain.

Let us consider each continuous interval in the domain separately:

In each case, let $\epsilon > 0$ be a small positive number.

Case (a) : $x < - 1$

For large negative values of $x$, $f \left(x\right)$ is small and positive.
At the other end of this interval, if $x = - 1 - \epsilon$ then

$f \left(x\right) = f \left(- 1 - \epsilon\right) \cong \frac{\sqrt{2}}{\left(\left(2 \times - 1\right) + 1\right) \left(- 1 - \epsilon + 1\right)}$

$= \frac{\sqrt{2}}{\epsilon} \to + \infty$ as $\epsilon \to 0$

So for $x < - 1$ the range of $f \left(x\right)$ is $\left(0 , + \infty\right)$

Case (b) : $- \frac{1}{2} < x \le 1$

f(-1/2+epsilon) ~= sqrt(3/2)//((2(-1/2+epsilon) + 1)(-1/2+1)

$= \frac{\sqrt{\frac{3}{2}}}{\epsilon} \to + \infty$ as $\epsilon \to 0$

$f \left(1\right) = \frac{0}{1} = 0$

So for $- \frac{1}{2} < x \le 1$ the range of $f \left(x\right)$ is $\left[0 , + \infty\right)$

Case (c) : $- 1 < x < - \frac{1}{2}$

$f \left(- 1 + \epsilon\right) \cong \frac{\sqrt{2}}{\left(\left(2 \times - 1\right) + 1\right) \left(- 1 + \epsilon + 1\right)}$

$= - \frac{\sqrt{2}}{\epsilon} \to - \infty$ as $\epsilon \to 0$

f(-1/2-epsilon) ~= sqrt(3/2)/((2(-1/2-epsilon) + 1)(-1/2+1)

$= - \frac{\sqrt{\frac{3}{2}}}{\epsilon} \to - \infty$ as $\epsilon \to 0$

So the interesting question is what is the maximum value of $f \left(x\right)$ in this interval. To find the value of $x$ for which this occurs look for the derivative to be zero.

$\frac{d}{\mathrm{dx}} f \left(x\right)$
$= \frac{\frac{1}{2} {\left(1 - x\right)}^{- \frac{1}{2}} \times - 1}{2 {x}^{2} + 3 x + 1} + \left({\left(1 - x\right)}^{\frac{1}{2}} \times - 1 \times {\left(2 {x}^{2} + 3 x + 1\right)}^{- 2} \times \left(4 x + 3\right)\right)$

$= \frac{- \frac{1}{2} {\left(1 - x\right)}^{- \frac{1}{2}}}{2 {x}^{2} + 3 x + 1} - \frac{{\left(1 - x\right)}^{\frac{1}{2}} \left(4 x + 3\right)}{2 {x}^{2} + 3 x + 1} ^ 2$

$= \frac{\left(- \frac{1}{2} {\left(1 - x\right)}^{- \frac{1}{2}} \left(2 {x}^{2} + 3 x + 1\right)\right) - \left({\left(1 - x\right)}^{\frac{1}{2}} \left(4 x + 3\right)\right)}{2 {x}^{2} + 3 x + 1} ^ 2$

This will be zero when the numerator is zero, so we would like to solve:

$- \frac{1}{2} {\left(1 - x\right)}^{- \frac{1}{2}} \left(2 {x}^{2} + 3 x + 1\right) - \left({\left(1 - x\right)}^{\frac{1}{2}} \left(4 x + 3\right)\right) = 0$

Multiply through by $2 {\left(1 - x\right)}^{\frac{1}{2}}$ to get:

$- \left(2 {x}^{2} + 3 x + 1\right) - 2 \left(1 - x\right) \left(4 x + 3\right) = 0$

That is:

$6 {x}^{2} - 5 x - 7 = 0$

which has roots $\frac{5 \pm \sqrt{25 + 4 \times 6 \times 7}}{12} = \frac{5 \pm \sqrt{194}}{12}$

Of these roots, $x = \frac{5 - \sqrt{194}}{12}$ falls in the interval concerned.

Substitute this back into $f \left(x\right)$ to find the maximum of #f(x) in this interval (approximately -10).

This seems over complex to me. Have I made any errors?

May 23, 2015

Answer: The range of the function is $\left(- \infty , - 10.58\right] \cup \left[0 , \infty\right)$

For $x \in \left(- \infty , - 1\right)$ $\to$ $y \in \left(0 , \infty\right)$
For $x \in \left(- 1 , - 0.5\right)$ $\to$ $y \in \left(- \infty , - 10.58\right]$
For $x \in \left(- 0.5 , 1\right]$ $\to$ $y \in \left[0 , \infty\right)$