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 ) ) }

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=(-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"].