Graphing a function - GMAT Quantitative

Card 0 of 56

Question

What is the domain of $y = 4 - x^{2}$ ?

Answer

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.

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Question

What is the domain of $y=-2\sqrt{x}$ ?

Answer

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 $x\geq 0$ .

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Question

What is the domain of the function $y=\sqrt{4-x^{2}}$ ?

Answer

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 $\left [ -2,2 \right ]$ .

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Question

Define the functions and as follows:

$\small \small f (x) = 2x + \sqrt{x - 1}$

$\small g (x) = 6x - \sqrt{x-1}$

What is the domain of the function ?

Answer

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,

$\small \small x - 1\geq 0$ , or

$\small \small \small x \geq 1$

Since the domains of and are the same, the domain of is also the same. In interval form the domain of is $\small \small [1, \infty )$

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Question

Define $f (x) = \frac{1}{x ^{2}- 25 }$ .

What is the natural domain of ?

Answer

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:

$x ^{2}-25 =0$

$x ^{2}= 25$

$x = -5 \textrm{ or }x = 5$

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

$(-\infty , -5 ) \cup \left ( -5,5\right ) \cup (5, \infty)$ .

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Question

Define $g (x) = \frac{1}{\sqrt[3]{x -27}}$

What is the natural domain of ?

Answer

The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression $\sqrt[3]{x -27}$ is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which

$\sqrt[3]{x -27} = 0$

$\left (\sqrt[3]{x -27} \right )^{3}= 0^{3}$

27 is the only number excluded from the domain.

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Question

Define $g(x) = \frac{x}{x^{2}+3x-4 }$

What is the natural domain of ?

Answer

Since the expression $x^{2}+3x-4$ is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which $x^{2}+3x-4 =0$ . We solve for by factoring the polynomial, which we can do as follows:

$x^{2}+3x-4 =0$

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

$x + 4 = 0 \Rightarrow x = -4$

$x -1 = 0 \Rightarrow x = 1$

Therefore, the domain excludes these two values of .

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Question

What is the domain of $y = 4 - x^{2}$ ?

Answer

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.

Compare your answer with the correct one above

Question

What is the domain of $y=-2\sqrt{x}$ ?

Answer

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 $x\geq 0$ .

Compare your answer with the correct one above

Question

What is the domain of the function $y=\sqrt{4-x^{2}}$ ?

Answer

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 $\left [ -2,2 \right ]$ .

Compare your answer with the correct one above

Question

Define the functions and as follows:

$\small \small f (x) = 2x + \sqrt{x - 1}$

$\small g (x) = 6x - \sqrt{x-1}$

What is the domain of the function ?

Answer

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,

$\small \small x - 1\geq 0$ , or

$\small \small \small x \geq 1$

Since the domains of and are the same, the domain of is also the same. In interval form the domain of is $\small \small [1, \infty )$

Compare your answer with the correct one above

Question

Define $f (x) = \frac{1}{x ^{2}- 25 }$ .

What is the natural domain of ?

Answer

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:

$x ^{2}-25 =0$

$x ^{2}= 25$

$x = -5 \textrm{ or }x = 5$

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

$(-\infty , -5 ) \cup \left ( -5,5\right ) \cup (5, \infty)$ .

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Question

Define $g (x) = \frac{1}{\sqrt[3]{x -27}}$

What is the natural domain of ?

Answer

The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression $\sqrt[3]{x -27}$ is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which

$\sqrt[3]{x -27} = 0$

$\left (\sqrt[3]{x -27} \right )^{3}= 0^{3}$

27 is the only number excluded from the domain.

Compare your answer with the correct one above

Question

Define $g(x) = \frac{x}{x^{2}+3x-4 }$

What is the natural domain of ?

Answer

Since the expression $x^{2}+3x-4$ is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which $x^{2}+3x-4 =0$ . We solve for by factoring the polynomial, which we can do as follows:

$x^{2}+3x-4 =0$

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

$x + 4 = 0 \Rightarrow x = -4$

$x -1 = 0 \Rightarrow x = 1$

Therefore, the domain excludes these two values of .

Compare your answer with the correct one above

Question

What is the domain of $y = 4 - x^{2}$ ?

Answer

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.

Compare your answer with the correct one above

Question

What is the domain of $y=-2\sqrt{x}$ ?

Answer

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 $x\geq 0$ .

Compare your answer with the correct one above

Question

What is the domain of the function $y=\sqrt{4-x^{2}}$ ?

Answer

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 $\left [ -2,2 \right ]$ .

Compare your answer with the correct one above

Question

Define the functions and as follows:

$\small \small f (x) = 2x + \sqrt{x - 1}$

$\small g (x) = 6x - \sqrt{x-1}$

What is the domain of the function ?

Answer

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,

$\small \small x - 1\geq 0$ , or

$\small \small \small x \geq 1$

Since the domains of and are the same, the domain of is also the same. In interval form the domain of is $\small \small [1, \infty )$

Compare your answer with the correct one above

Question

Define $f (x) = \frac{1}{x ^{2}- 25 }$ .

What is the natural domain of ?

Answer

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:

$x ^{2}-25 =0$

$x ^{2}= 25$

$x = -5 \textrm{ or }x = 5$

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

$(-\infty , -5 ) \cup \left ( -5,5\right ) \cup (5, \infty)$ .

Compare your answer with the correct one above

Question

Define $g (x) = \frac{1}{\sqrt[3]{x -27}}$

What is the natural domain of ?

Answer

The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression $\sqrt[3]{x -27}$ is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which

$\sqrt[3]{x -27} = 0$

$\left (\sqrt[3]{x -27} \right )^{3}= 0^{3}$

27 is the only number excluded from the domain.

Compare your answer with the correct one above

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