Find range of function $f(x)=\frac3{(2-x²)}$

Question

My approach -

let $$ f(x)=yx^2 - 2y + 3 = 0 \qquad x= \sqrt{\frac{2y-3}{y}} $$

This implies that $(2y+3)/y\ge 0$ and $y\ne 0$. Hence $y=[3/2, \infty)$

But my textbook says $y = (-\infty, 0) \cup [3/2, \infty)$ .

Answer 1

You should not take the square root. Instead, write it as

$$x^2=\frac{2y-3}{y}\ge0$$

Then, consider two cases:

1) $2y-3\ge0$ and $y>0$, which leads to $y\ge \frac32$

2) $2y-3<0$ and $y<0$, which leads to $y<0$

As a result, $y = (-\infty, 0) \cup [3/2, \infty)$

Answer 2

$(2y-3)/y\ge 0$, yes. But if $y$ is negative, you have to change the sense of the inequality when you divide by it (see for instance this). So if $y<0$, we get $2y-3\le 0$.

Answer 3

$y=\frac{3}{2-x^2}, x \in R-\{\pm \sqrt{2}\}$ This function has min at $x=0$ as $f'(0)=-3(2-x^2)^{-2}(-2x)=0$ but it can attain any large positive or large negative values near $x=\pm \sqrt{2}$ So the range of $f(x)$ is all real numbers.

Answer 4

The sign of $\dfrac{2y-3}y$, when $y\ne 0$, is the sign of the quadratic polynomial $y(2y-3)$, and by a basic high-school result, it is positive outside the interval of the roots $\bigl[0,\frac32\bigr]$.