State the domain and range for a given function

[brotherbobby]

Homework Statement

State and domain and range for the following function : $g(x) = \dfrac{1}{1+x^2}$

Relevant Equations

The domain of a function, which I'd write as $\mathscr{D}\{f(x)\}$ are all acceptable values that $x$ can take so that the function is defined. The range of a function, $\mathscr{R}\{f(x)\}$ are all values of the function for the values of $x$ in the domain.

Attempt : Let me copy and paste the problem as it appeared in the text. Please note that the given problem appears in part (b), which I have underlined in red ink in this way $\color{red}{\rule{50pt}{1pt}}$

Clearly the domain is $\boxed{\mathscr{D}\{f(x)\}: x\in\mathbb{R}}\quad\color{green}{\Large{\checkmark}}$
Shown is the answer in the text (part (b)).

The range can be calculated in a number of ways. It is the third way with which I have a problem and need your input.

$\rm{I :}$ $\textsf{Here I proceed towards the function incrementally without finding its inverse}$
\begin{align*}
-\infty &< x< \infty \\
\Rightarrow 0 &\le x^2 < \infty \\
\Rightarrow 1&\le1+x^2 < \infty \\
\Rightarrow 0&<\dfrac{1}{1+x^2}\le 1\\
\end{align*} Hence the range of the function $\boxed{\mathscr{R}\{f(x)\}\in (0,1]}\quad\color{green}{\Large{\checkmark}}$. The answer in the text (shown above) is wrong to include the point $0$, or it is me mistaken?

$\rm{II :}$ $\textsf{Here I find the inverse of the function first and then try to find the domain of the inverse, or y}$

If $y=\dfrac{1}{1+x^2}\Rightarrow 1+x^2=\dfrac{1}{y}\Rightarrow x^2=\dfrac{1-y}{y}\Rightarrow x=\sqrt{\dfrac{1-y}{y}}$
Clearly, $y\ne 0$. Also the argument $\dfrac{1-y}{y}\ge0$. This can mean either (a) $1-y\ge0$ and $y>0$ which would lead to $y\le1$ and $y>0$, which leads to the correct range found earlier : $\boxed{\mathscr{R}\{f(x)\}\in (0,1]}\quad\color{green}{\Large{\checkmark}}$. Or it could also mean (b) $1-y\le 0\Rightarrow y\ge 1$ and $y<0$, which is not possible.

$\rm{III :}$ $\textsf{A variant of the above method where I find the inverse again, but leads me to trouble}$

If $y=\dfrac{1}{1+x^2}\Rightarrow 1+x^2=\dfrac{1}{y}\Rightarrow x^2=\dfrac{1}{y}-1\Rightarrow x=\sqrt{\dfrac{1}{y}-1}$. Now $\dfrac{1}{y}-1\ge 0\Rightarrow \dfrac{1}{y}\ge 1\Rightarrow y\le 1$. Hence, the range comes out to be $\boxed{\mathscr{R}\{f(x)\}\in (-\infty,1]}\quad\color{red}{\huge{\times}}$

Request : Where did I go wrong in $\textrm{III}$? Many thanks.

 

[Orodruin]

$1/y \geq 1$ does not imply only $y \leq 1$. Consider $y = -1$ as a counter example.

 

[pasmith]

III: The condition is $\frac 1y \geq 1$. Now there are two cases: if $y > 0$ then multpliying by $y$ preserves the inequality, so $1 \geq y$. Alternatively, if $y < 0$ then muitpliying by $y$ reverses the inequality, so that $1 \leq y$. This is impossible, since $y < 0 < 1$. You can also see directly that if $y < 0$ then $y^{-1}+ < 0$ and $y^{-1} -1 < 0$.

 

[brotherbobby]

$1/y \geq 1$ does not imply only $y \leq 1$. Consider $y = -1$ as a counter example.

Indeed. Let's take $y=-2$ as a counterexample - there are too many $1$'s here.
$y=-2<1\nRightarrow \dfrac{1}{y}>1$. (Of course $-\dfrac{1}{2}\ngtr 1)$

 

[brotherbobby]

You can also see directly that if y<0 then $y^{-1+ < 0$ and y−1−1<0.

I didn't get the last part of your reply @pasmith

 

[FactChecker]

Hence the range of the function $\boxed{\mathscr{R}\{f(x)\}\in (0,1]}\quad\color{green}{\Large{\checkmark}}$. The answer in the text (shown above) is wrong to include the point $0$, or it is me mistaken?

I agree with you.

$\rm{II :}$ $\textsf{Here I find the inverse of the function first and then try to find the domain of the inverse, or y}$

If $y=\dfrac{1}{1+x^2}\Rightarrow 1+x^2=\dfrac{1}{y}\Rightarrow x^2=\dfrac{1-y}{y}\Rightarrow x=\sqrt{\dfrac{1-y}{y}}$

This is treacherous. It is safer to do this in several steps, indicating the values of y that are valid for each step.

Clearly, $y\ne 0$. Also the argument $\dfrac{1-y}{y}\ge0$. This can mean either (a) $1-y\ge0$ and $y>0$ which would lead to $y\le1$ and $y>0$, which leads to the correct range found earlier : $\boxed{\mathscr{R}\{f(x)\}\in (0,1]}\quad\color{green}{\Large{\checkmark}}$. Or it could also mean (b) $1-y\le 0\Rightarrow y\ge 1$ and $y<0$, which is not possible.

I agree.

$\rm{III :}$ $\textsf{A variant of the above method where I find the inverse again, but leads me to trouble}$

If $y=\dfrac{1}{1+x^2}\Rightarrow 1+x^2=\dfrac{1}{y}\Rightarrow x^2=\dfrac{1}{y}-1\Rightarrow x=\sqrt{\dfrac{1}{y}-1}$. Now $\dfrac{1}{y}-1\ge 0\Rightarrow \dfrac{1}{y}\ge 1\Rightarrow y\le 1$. Hence, the range comes out to be $\boxed{\mathscr{R}\{f(x)\}\in (-\infty,1]}\quad\color{red}{\huge{\times}}$

Request : Where did I go wrong in $III$? Many thanks.

1) When you divide by $y$, you should indicate that $y \ne 0$. This is a problem when you chain a bunch of steps together in one line and do not keep track of what values are illegal at each step.
2) The square root symbol, $\sqrt a$ only represents the positive result of the square root. When you introduce it, you should indicate both positive and negative values, with $\pm \sqrt a$, unless you state some reason to omit one or the other.

 

[Mark44]

2) The square root symbol $\sqrt a$ only represents the positive result of the square root. When you introduce it, you should indicate both positive and negative values, with $\pm \sqrt a$, unless you state some reason to omit one or the other.

To elaborate, for the benefit of the OP, taking the square root of both sides of an equation can produce incorrect or partially correct results if you're not careful.
For example, starting with $x^2 = 9$ it is intuitive, but incorrect to conclude that $x = 3$.

One way to look at it is this: $x^2 = 9 \Rightarrow \sqrt{x^2} = \sqrt 9 \Rightarrow |x| = 3$. This means that x = 3 or x = -3; i.e., that $x = \pm 3$.

Note that $\sqrt{x^2} \ne x$. A counterexample is $\sqrt{(-3)^2} = 3 \ne -3$.

 

[Orodruin]

I didn't get the last part of your reply @pasmith

There is an extra + that should not be there:

$y^{-1}+ < 0$

should be $y^{-1} < 0$, which indeed implies that $y^{-1} < 0 < 1$.