randRange( 1, 10 ) randRange( 1, 10 ) randRange( 1, 10 ) function() { return "\\{ \\, x \\in \\RR \\mid " + Array.prototype.join.call( arguments, ", \\," ) + "\\, \\}"; } function( n, sym ) { return "x " + sym + n; } function( n, m, sym1, sym2 ) { return n + sym1 + " x " + sym2 + m; } function( n ) { return FN( n, "\\neq" ); } function( n ) { return FN( n, "\\geq" ); } function( n ) { return FN( n, "\\leq" ); } function( n ) { return FN( n, ">" ); } function( n ) { return FN( n, "<" ); } function( n, m ) { return FN2( n, m, "<", "\\leq" ); } function( n, m ) { return FN2( n, m, "\\leq", "\\leq" ); } function( n, m ) { return FN2( n, m, "<", "<" ); } { "two-denom-simplify": SET( NEQ( -1*A ), NEQ( B ) ), "two-denom-cond": SET( NEQ( -1*A ) ), "sqrt": SET( GEQ( A ) ), "inverse-sqrt": SET( GE( A ) ), "inverse-sqrt-cond": SET( NEQ( A ) ), "sqrt-frac": SET( LE_LEQ( A, A+B ) ), "two-denom-cond-weird": SET( NEQ( -1*A ), NEQ( C ) ), "sqrt-poly-frac": SET( GEQ( C ) ), "sqrt-abs": SET( LEQ_LEQ( -1*A, A ) ), "inverse-sqrt-abs": SET( LE_LE( -1*A, A ) ) } $._("if")

`f(x) = \dfrac{ x + A }{ ( x + A )( x - B ) }`

What is the domain of the real-valued function `f(x)`?

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

- `c`

`f(x)` is undefined when the denominator is 0.

The denominator is 0 when `x=(-1*A)` or `x=B`.

So we know that `x \neq -1*A` and `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 + A }{ ( x + A )( x - B ) }`, is undefined when its denominator is 0.

The denominator is 0 when `x=-1*A` or `x=B`.

So, based on the first piecewise definition, we know that `x \neq -1*A` and `x \neq B`.

However, the second piecewise definition applies when `x = B`, and the second piecewise definition, `C`, has no weird gaps or holes, so `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 - A`, must be greater than or equal to zero.

So `x - A \geq 0`; this means `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 - A`, must be greater than or equal to zero.

So `x - A \geq 0`; this means `x \geq A`.

Next, we also need to consider that `f(x)` is undefined when the denominator, `\sqrt{ x - A }`, is zero.

So `\sqrt{ x - A } \neq 0`.

`\sqrt{ z } = 0` only when `z = 0`, so `\sqrt{ x - A } \neq 0` means that `x - A \neq 0`.

So `x \neq A`.

So we have two restrictions: `x \geq A` and `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 - A } }`, is undefined where the denominator is zero and where the radicand (the expression under the radical) is less than zero.

The denominator, `\sqrt{ x - A }`, is zero when `x - A = 0`, so we know that `x \neq A`.

The radicand, `x - A`, is less than zero when `x < A`, so we know that `x \geq A`.

So the first piecewise definition of `f(x)` is defined when `x \neq A` and `x \geq A`. Combining these two restrictions, the first piecewise definition is defined when `x > A`. The first piecewise defintion applies when `x \geq A`, so this restriction is relevant.

The second piecewise definition of `f(x)`, `\frac{ 1 }{ \sqrt{ A - x } }`, applies when `x < A` and is undefined where the denominator is zero and where the radicand is less than zero.

The denominator, `\sqrt{ A - x }`, is zero when `A - x = 0`, so we know that `x \neq A`.

The radicand, `A - x`, is less than zero when `x > A`, so we know that `x \leq A`.

So the second piecewise definition of `f(x)` is defined when `x \neq A` and `x \leq A`. Combining these two restrictions, the second piecewise definition is defined when `x < A`. However, the second piecewise definition of `f(x)` only applies when `x < A`, so restriction isn't actually relevant to the domain of `f(x)`.

So the first piecewise definition is defined when `x > A` and applies when `x \geq A`; the second piecewise definition is defined when `x < A` and applies when `x < A`. Putting the restrictions of these two together, the only place where a definition applies and the value is undefined is at `x = A`. So the only restriction on the domain of `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 > A+B`, so we know `x \leq A+B`.

The bottom radical is undefined when `x - A < 0`.

So the bottom radical is undefined when `x < A`, so we know `x \geq A`.

Next, we need to consider that `f(x)` is undefined anywhere where the denominator, `\sqrt{ x - A }`, is zero.

So `\sqrt{ x - A } \neq 0`, so `x - A \neq 0`, so `x \neq A`.

So we have three restrictions: `x \leq A+B`, `x \geq A`, and `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 + A }{ ( x + A )( x - C ) }`, is undefined where the denominator is zero.

The denominator, `(x + A)(x - C)`, is zero when `x = -1*A` or `x = C`.

So the first piecewise definition of `f(x)` is defined when `x \neq -1*A` and `x \neq C`. The first piecewise definition applies when `x = -1*A` and `x = C`, so these restrictions are relevant to the domain of `f(x)`.

The second piecewise definition of `f(x)`, `A`, is a simple horizontal line function and so has no holes or gaps to worry about, so it is defined everywhere.

So the first piecewise definition is defined when `x \neq -1*A` and `x \neq C` and applies when `x \neq B`; the second piecewise definition is defined always and applies when `x = B`. Putting the restrictions of these two together, the only places where a definition applies and is undefined are at `x = -1*A` and `x = C`. So the restriction on the domain of `f(x)` is that `x \neq -1*A` and `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 - C \geq 0`, which means `x \geq C`.

Next, we also need to consider that `f(x)` is undefined anywhere where the denominator is zero.

So `x \neq -1*A` and `x \neq -1*B`.

However, these last two restrictions are irrelevant since `C > -1*A` and `C > -1*B` and so `x \geq C` will ensure that `x \neq -1*A` and `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 A` and `x \geq -1*A`; or, equivalently, `-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 A`, which means `-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{ A - \lvert x \rvert } \neq 0`, so `\lvert x \rvert \neq A`.

This means that `x \neq A` and `x \neq -1*A`.

So we have four restrictions: `x \geq -1*A`, `x \leq A`, `x \neq -1*A`, and `x \neq A`.

Combining these four, we know that `x > -1*A` and `x < A`; alternatively, that `-1*A < x < A`.

Expressing this mathematically, the domain is `CHOICES["inverse-sqrt-abs"]`.