`f(x) = \dfrac{ x + `

`A` }{ ( x + `A` )( x - `B` ) }

What is the domain of `f(x)`

?

`CHOICES["two-denom-simplify"]`

`c`

`f(x)`

is undefined when the denominator is 0.

The denominator is 0 when `x=`

or `(-1*A)``x=`

.`B`

So we know that `x \neq `

and `-1*A``x \neq `

.`B`

Expressing this mathematically, the domain is

.`CHOICES["two-denom-simplify"]`

`f(x)= \begin{cases} \dfrac{ x + `

`A` }{ ( x + `A` )( x - `B` ) } & \text{if $x \neq `B`$} \\ `C` & \text{if $x = `B`$} \end{cases}

`CHOICES["two-denom-cond"]`

`c`

`f(x)`

is a piecewise function, so we need to examine where each piece is undefined.

The first piecewise definition of `f(x)`

, `\frac{ x + `

, is undefined when its denominator is 0.`A` }{ ( x + `A` )( x - `B` ) }

The denominator is 0 when `x=`

or `-1*A``x=`

.`B`

So, based on the first piecewise definition, we know that `x \neq `

and `-1*A``x \neq `

.`B`

However, the second piecewise definition applies when `x = `

, and the second piecewise definition, `B`

, has no weird gaps or holes, so `C``f(x)`

is defined at `x = `

.`B`

So the only restriction on the domain is that `x \neq `

.`-1*A`

Expressing this mathematically, the domain is

.`CHOICES["two-denom-cond"]`

`f(x) = \sqrt{ x - `

`A` }

`CHOICES["sqrt"]`

`c`

`f(x)`

is undefined when the radicand (the expression under the radical) is less than zero.

So the radicand, `x - `

, must be greater than or equal to zero.`A`

So `x - `

; this means `A` \geq 0`x \geq `

.`A`

Expressing this mathematically, the domain is

.`CHOICES["sqrt"]`

`f(x) = \dfrac{ 1 }{ \sqrt{ x - `

`A` } }

`CHOICES["inverse-sqrt"]`

`c`

First, we need to consider that `f(x)`

is undefined when the radicand (the expression under the radical) is less than zero.

So the radicand, `x - `

, must be greater than or equal to zero.`A`

So `x - `

; this means `A` \geq 0`x \geq `

.`A`

Next, we also need to consider that `f(x)`

is undefined when the denominator, `\sqrt{ x - `

, is zero.`A` }

So `\sqrt{ x - `

.`A` } \neq 0

`\sqrt{ z } = 0`

only when `z = 0`

, so `\sqrt{ x - `

means that `A` } \neq 0`x - `

.`A` \neq 0

So `x \neq `

.`A`

So we have two restrictions: `x \geq `

and `A``x \neq `

.`A`

Combining these two restrictions, we are left with simply `x > `

.`A`

Expressing this in mathematical notation, the domain is

.`CHOICES["inverse-sqrt"]`

`f(x) = \begin{cases} \dfrac{ 1 }{ \sqrt{ x - `

`A` } } & \text{if $x \geq `A`$} \\ \dfrac{ 1 }{ \sqrt{ `A` - x } } & \text{if $x < `A`$} \end{cases}

`CHOICES["inverse-sqrt-cond"]`

`c`

`f(x)`

is a piecewise function, so we need to examine where each piece is undefined.

The first piecewise definition of `f(x)`

, `\frac{ 1 }{ \sqrt{ x - `

, is undefined where the denominator is zero and where the radicand (the expression under the radical) is less than zero.`A` } }

The denominator, `\sqrt{ x - `

, is zero when `A` }`x - `

, so we know that `A` = 0`x \neq `

.`A`

The radicand, `x - `

, is less than zero when `A``x < `

, so we know that `A``x \geq `

.`A`

So the first piecewise definition of `f(x)`

is defined when `x \neq `

and `A``x \geq `

. Combining these two restrictions, the first piecewise definition is defined when `A``x > `

. The first piecewise defintion applies when `A``x \geq `

, so this restriction is relevant.`A`

The second piecewise definition of `f(x)`

, `\frac{ 1 }{ \sqrt{ `

, applies when `A` - x } }`x < `

and is undefined where the denominator is zero and where the radicand is less than zero.`A`

The denominator, `\sqrt{ `

, is zero when `A` - x }

, so we know that `A` - x = 0`x \neq `

.`A`

The radicand,

, is less than zero when `A` - x`x > `

, so we know that `A``x \leq `

.`A`

So the second piecewise definition of `f(x)`

is defined when `x \neq `

and `A``x \leq `

. Combining these two restrictions, the second piecewise definition is defined when `A``x < `

. However, the second piecewise definition of `A``f(x)`

only applies when `x < `

, so restriction isn't actually relevant to the domain of `A``f(x)`

.

So the first piecewise definition is defined when `x > `

and applies when `A``x \geq `

; the second piecewise definition is defined when `A``x < `

and applies when `A``x < `

. Putting the restrictions of these two together, the only place where a definition applies and the value is undefined is at `A``x = `

. So the only restriction on the domain of `A``f(x)`

is `x \neq `

.`A`

Expressing this mathematically, the domain is

.`CHOICES["inverse-sqrt-cond"]`

`f(x) = \dfrac{ \sqrt{ `

`A+B` - x } }{ \sqrt{ x - `A` } }

`CHOICES["sqrt-frac"]`

`c`

First, we need to consider that `f(x)`

is undefined anywhere where either radical is undefined, so anywhere where either radicand (the expression under the radical symbol) is less than zero.

The top radical is undefined when

.`A+B` - x < 0

So the top radical is undefined when `x > `

, so we know `A+B``x \leq `

.`A+B`

The bottom radical is undefined when `x - `

.`A` < 0

So the bottom radical is undefined when `x < `

, so we know `A``x \geq `

.`A`

Next, we need to consider that `f(x)`

is undefined anywhere where the denominator, `\sqrt{ x - `

, is zero.`A` }

So `\sqrt{ x - `

, so `A` } \neq 0`x - `

, so `A` \neq 0`x \neq `

.`A`

So we have three restrictions: `x \leq `

, `A+B``x \geq `

, and `A``x \neq `

.`A`

Combining these three restrictions, we know that

.`A` < x \leq `A+B`

Expressing this mathematically, the domain is

.`CHOICES["sqrt-frac"]`

`f(x) = \begin{cases} \dfrac{ x + `

`A` }{ ( x + `A` )( x - `C` ) } & \text{if $x \neq `B`$} \\ `A` & \text{if $x = `B`$} \end{cases}

`CHOICES["two-denom-cond-weird"]`

`c`

`f(x)`

is a piecewise function, so we need to examine where each piece is undefined.

The first piecewise definition of `f(x)`

, `\frac{ x + `

, is undefined where the denominator is zero.`A` }{ ( x + `A` )( x - `C` ) }

The denominator, `(x + `

, is zero when `A`)(x - `C`)`x = `

or `-1*A``x = `

.`C`

So the first piecewise definition of `f(x)`

is defined when `x \neq `

and `-1*A``x \neq `

. The first piecewise definition applies when `C``x = `

and `-1*A``x = `

, so these restrictions are relevant to the domain of `C``f(x)`

.

The second piecewise definition of `f(x)`

,

, is a simple horizontal line function and so has no holes or gaps to worry about, so it is defined everywhere.`A`

So the first piecewise definition is defined when `x \neq `

and `-1*A``x \neq `

and applies when `C``x \neq `

; the second piecewise definition is defined always and applies when `B``x = `

. Putting the restrictions of these two together, the only places where a definition applies and is undefined are at `B``x = `

and `-1*A``x = `

. So the restriction on the domain of `C``f(x)`

is that `x \neq `

and `-1*A``x \neq `

.`C`

Expressing this mathematically, the domain is

.`CHOICES["two-denom-cond-weird"]`

`f(x) = \dfrac{ \sqrt{ x - `

`C` } }{ x^2 + `A+B` x + `A*B` }

`CHOICES["sqrt-poly-frac"]`

`c`

`f(x) = \dfrac{ \sqrt{ x - `

`C` } }{ x^2 + `A+B` x + `A*B` } = \dfrac{ \sqrt{ x - `C` } }{ ( x + `A` )( x + `B` ) }

First, we need to consider that `f(x)`

is undefined anywhere where the radical is undefined, so the radicand (the expression under the radical) cannot be less than zero.

So `x - `

, which means `C` \geq 0`x \geq `

.`C`

Next, we also need to consider that `f(x)`

is undefined anywhere where the denominator is zero.

So `x \neq `

and `-1*A``x \neq `

.`-1*B`

However, these last two restrictions are irrelevant since

and `C` > `-1*A`

and so `C` > `-1*B``x \geq `

will ensure that `C``x \neq `

and `-1*A``x \neq `

.`-1*B`

Combining these restrictions, then, leaves us with simply `x \geq `

.`C`

Expressing this mathematically, the domain is

.`CHOICES["sqrt-poly-frac"]`

`f(x) = \sqrt{ `

`A` - \lvert x \rvert }

`CHOICES["sqrt-abs"]`

`c`

`f(x)`

is undefined when the radicand (the expression under the radical) is less than zero.

So we know that

.`A` - \lvert x \rvert \geq 0

So `\lvert x \rvert \leq `

.`A`

This means `x \leq `

and `A``x \geq `

; or, equivalently, `-1*A`

.`-1*A` \leq x \leq `A`

Expressing this mathematically, the domain is

.`CHOICES["sqrt-abs"]`

`f(x) = \dfrac{ `

`B` }{ \sqrt{ `A` - \lvert x \rvert } }

`CHOICES["inverse-sqrt-abs"]`

`c`

First, we need to consider that `f(x)`

is undefined anywhere where the radicand (the expression under the radical) is less than zero.

So we know that

.`A` - \lvert x \rvert \geq 0

This means `\lvert x \rvert \leq `

, which means `A`

.`-1*A` \leq x \leq `A`

Next, we need to consider that `f(x)`

is also undefined anywhere where the denominator is zero.

So we know that `\sqrt{ `

, so `A` - \lvert x \rvert } \neq 0`\lvert x \rvert \neq `

.`A`

This means that `x \neq `

and `A``x \neq `

.`-1*A`

So we have four restrictions: `x \geq `

, `-1*A``x \leq `

, `A``x \neq `

, and `-1*A``x \neq `

.`A`

Combining these four, we know that `x > `

and `-1*A``x < `

; alternatively, that `A`

.`-1*A` < x < `A`

Expressing this mathematically, the domain is

.`CHOICES["inverse-sqrt-abs"]`