Graphing a function - GMAT Quantitative

To find the domain, we need to decide which values can take. The is under a square root sign, so cannot be negative. can, however, be 0, because we can take the square root of zero. Therefore the domain is xgeq 0.

To find the domain, we must find the interval on which $$\sqrt{4-x^{2}$$} is defined. We know that the expression under the radical must be positive or 0, so $$\sqrt{4-x^{2}$$} is defined when $x^{2}$leq 4. This occurs when x geq -2 and x leq 2. In interval notation, the domain is .

The domain of is the intersection of the domains of and . and are each restricted to all values of that allow the radicand to be nonnegative - that is,

, or

Since the domains of and are the same, the domain of is also the same. In interval form the domain of is

The only restriction on the domain of is that the denominator cannot be 0. We set the denominator to 0 and solve for to find the excluded values:

The domain is the set of all real numbers except those two - that is,

.

The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which

27 is the only number excluded from the domain.

Since the expression is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which . We solve for by factoring the polynomial, which we can do as follows:

Replacing the question marks with integers whose product is and whose sum is 3:

Therefore, the domain excludes these two values of .

The domain of the function specifies the values that can take. Here, $4-x^{2}$ is defined for every value of , so the domain is all real numbers.

To find the domain, we need to decide which values can take. The is under a square root sign, so cannot be negative. can, however, be 0, because we can take the square root of zero. Therefore the domain is xgeq 0.

To find the domain, we must find the interval on which $$\sqrt{4-x^{2}$$} is defined. We know that the expression under the radical must be positive or 0, so $$\sqrt{4-x^{2}$$} is defined when $x^{2}$leq 4. This occurs when x geq -2 and x leq 2. In interval notation, the domain is .

The domain of is the intersection of the domains of and . and are each restricted to all values of that allow the radicand to be nonnegative - that is,

, or

Since the domains of and are the same, the domain of is also the same. In interval form the domain of is

The only restriction on the domain of is that the denominator cannot be 0. We set the denominator to 0 and solve for to find the excluded values:

The domain is the set of all real numbers except those two - that is,

.

The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which

27 is the only number excluded from the domain.

Since the expression is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which . We solve for by factoring the polynomial, which we can do as follows:

Replacing the question marks with integers whose product is and whose sum is 3:

Therefore, the domain excludes these two values of .

The domain of the function specifies the values that can take. Here, $4-x^{2}$ is defined for every value of , so the domain is all real numbers.

To find the domain, we need to decide which values can take. The is under a square root sign, so cannot be negative. can, however, be 0, because we can take the square root of zero. Therefore the domain is xgeq 0.

To find the domain, we must find the interval on which $$\sqrt{4-x^{2}$$} is defined. We know that the expression under the radical must be positive or 0, so $$\sqrt{4-x^{2}$$} is defined when $x^{2}$leq 4. This occurs when x geq -2 and x leq 2. In interval notation, the domain is .

The domain of is the intersection of the domains of and . and are each restricted to all values of that allow the radicand to be nonnegative - that is,

, or

Since the domains of and are the same, the domain of is also the same. In interval form the domain of is

The only restriction on the domain of is that the denominator cannot be 0. We set the denominator to 0 and solve for to find the excluded values:

The domain is the set of all real numbers except those two - that is,

.

The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which

27 is the only number excluded from the domain.