Range of rational function in different ways.

Question

I was looking for range of a rational function, I used three ways to get the answer and all of them are giving different results . e.g. f(x)=1/(1-x²) then,

(1)

 let 1/(1-x²) =y, (1-x²)=1/y
or
x²= 1-(1/y) or x= ±{1-(1/y)}½
Now 1-1/y must be ≥0
Therefore 1-1/y≥0 or 1≥1/y or y≥1
Therefore range (f)={1, ♾️}

(2)

let 1/(1-x²)=y then, 1= y-yx²
or (y)x²+0.x+(1-y)=0
Clearly D≥0 therefore -4y(1-y)≥0
Or
(i)
1-y≤0 or 1≤y 

this means range(f)={1,♾️}

(ii)

    -4y+4y²≥0 or y²≥y or y≥1
Again result is same as 2(i)

(3)

in f(x)= 1/(1-x²), putting
x= 0.0000001, y≈1
x≈ 0.999...., y≈ ♾️
x= 1.000...1, y≈ - ♾️
x= ±1000...,  y≈ 0
Therefore
range (f)={(1, ♾️) U (-♾️, 0}

You can see that 1 and 2 yield same result but they are incorrect, I want to know that what is the mistake in logics there.

Note- I have rewritten the whole thing again after some of the comments below.

Answer 1

We wish to find the range of the function $f(x) = \dfrac{1}{1 - x^2}$.

In what follows, I will assume that the range is a subset of the real numbers and that the domain is the largest subset of the real numbers for which the function is defined. Since the function is defined for every real number except those that make the denominator equal to zero, the domain is the set of all real numbers except $1, -1$. Thus, $$\text{Dom}_f = (-\infty, -1) \cup (-1, 1) \cup (1, \infty)$$

Method 1: You set $$y = \frac{1}{1 - x^2}$$ then took reciprocals. Before we can do that, we must first establish that $y \neq 0$.

If $y = 0$, we obtain $$\frac{1}{1 - x^2} = 0$$ Multiplying both sides of the equation by $1 - x^2$ yields $$1 = 0$$ which is a contradiction. Therefore, $y \neq 0$, so we can take reciprocals whenever $y$ is in the range.

\begin{align*} y & = \frac{1}{1 - x^2}\\ \frac{1}{y} & = 1 - x^2\\ x^2 & = 1 - \frac{1}{y} \end{align*} Thus, we require that $1 - \dfrac{1}{y} \geq 0$.

Observe that when $y < 0$, then $1 - \dfrac{1}{y} > 1 > 0$. You overlooked this case.

If $y > 0$, then \begin{align*} 1 - \frac{1}{y} & \geq 0\\ 1 & \geq \frac{1}{y}\\ y & \geq 1 \end{align*} where we have multiplied both sides of the inequality by $y > 0$ in the final step. Since $y > 0$, the direction of the inequality is preserved.

Hence, the range is $$\text{Ran}_f = (-\infty, 0) \cup [1, \infty)$$

Method 2: We know that if $y$ is in the range, then there is a real number $x$ such that $y = f(x)$. \begin{align*} y & = \frac{1}{1 - x^2}\\ y - yx^2 & = 1\\ 0 & = yx^2 + 1 - y \end{align*} which is a quadratic equation in $x$ provided that $y \neq 0$.

If $y = 0$, we obtain $0 = 1$, which is a contradiction. Hence, $y \neq 0$.

Since a quadratic equation has real roots whenever its discriminant $\Delta = b^2 - 4ac \geq 0$, there exists a real number $x$ such that $f(x) = y$ whenever \begin{align*} \Delta & \geq 0\\ b^2 - 4ac & \geq 0\\ 0^2 - 4y(1 - y) & \geq 0\\ 4y(y - 1) & \geq 0 \end{align*} and $y \neq 0$. The inequality holds when $y = 1$ or when the terms $y$ and $y - 1$ have the same sign. They have the same sign when $y < 0$ or $y > 1$. Hence, the range of the function is $$\text{Ran}_f = (-\infty, 0) \cup \{1\} \cup (1, \infty) = (-\infty, 0) \cup [1, \infty)$$ as we found above.

In both your attempts at this method, it looks like you divided by $y$, which caused you lose information. Rather than dividing by $y$, you should have factored the expression, as shown above.

Answer 2

Method 1:

let 1/(1-x²) =y, (1-x²)=1/y

so you should note that $y \ne 0$.

or

x²= 1-(1/y) or x= ±{1-(1/y)}½

Now 1-1/y must be ≥0

Therefore 1-1/y≥0 or 1≥1/y ...

So far so good but the next step is wrong

or y≥1

$1\ge \frac 1y \not \implies y\ge 1$

$1\ge \frac 1y \implies \begin{cases} y\ge 1&y> 0\\y\le 1&y < 0\end{cases}$

Therefore range (f)={1, ♾️}

Only if we assume $y$ is positive. If $y$ is negative then we have the only restriction being that $y \le 1$ but as $y < 0 \le 1$ that is redundant.

So the range is $(-\infty, 0)\cup [1,\infty)$.

....

Method 2:

let 1/(1-x²)=y then, 1= y-yx²

or (y)x²+0.x+(1-y)=0

Clearly D≥0 therefore -4y(1-y)≥0

So this means you have one of the following two cases.

Either

ONE: $1-y\ge 0$ AND $-4y\ge 0$.

(i)

1-y≤0 or 1≤y

You transposed the inequality symbol. You should have written

$1-y \ge 0$ or $1 \ge y$.

But You forgot to include $-4y \ge 0$. That will affect your answer:

$1-y\ge 0$ AND $-4y\ge 0$.

So $1 \ge y$ and $y\le \frac 0{-4} = 0$.

So $y \le 1$ AND $y \le 0$ so $y\le 0$. Combining that is $y \le 0 \le 1$. So $y \le 0$.

But $y = 0$ is impossible so $y < 0$ is part of the range.

Or

TWO: $1-y \le 0$ AND $-4y \le 0$

so $1\le y$ and $y \ge \frac 0{-4} = 0$. So $y \ge 1$ and $y \ge 0$. Combining we have $y\ge 1$.

So $[1,\infty)$ is the other part of the range.

Putting part ONE and TWO together we have either $y < 0$ or $y \ge 1$ so the range is $(-\infty, 0)\cup [1,-\infty)$.

-4y+4y²≥0 or y²≥y or y≥1

Again result is same as 2(i)

$y^2 \ge y \not \implies y \ge 1$.

It means if $y > 0$ then $y \ge 1$ but if $y=0$ you can not divide and if $y < 0$ then the inequality "flips".

So we have either.

a) $y > 0$ and $y\ge 1$ or

b) $y < 0$ and $y \le 1$ or

c) $y = 0$ but this is not possible.

So for the third time we get that the range is

$(-infty, 0) \cup [1, \infty)$.

All three methods give the same results.