Domain and range of a function - HiSET

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Define functions $f(x) = $\sqrt{x}$$ and $g(x) =- $x^{2}$$

Give the domain of the composition $g\circ f$ .

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The domain of the composition of functions $g\circ f$ is that the set of all values that fall in the domain of such that falls in the domain of .

$f(x) = $\sqrt{x}$$ , a square root function, whose domain is the set of all values of that make the radicand nonnegative. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

$g(x) =- $x^{2}$$ , a polynomial function, whose domain is the set of all real numbers. Therefore, the domain is not restricted further by . The domain of $g\circ f$ is $\left [ 0, \infty\right )$ .

Define functions $f(x) = $\frac{1}{x-1}$$ and $g(x)= $\sqrt{x}$+ 1$ .

Give the domain of the composition $g\circ f$ .

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The domain of the composition of functions $g\circ f$ is that the set of all values that fall in the domain of such that falls in the domain of .

$f(x) = $\frac{1}{x-1}$$ , a rational function, so its domain is the set of all numbers except those that make its denominator zero. if and only if , making its domain $(-\infty, 1) \cup (1, \infty)$ .

$g(x)= $\sqrt{x}$+ 1$ , a square root function, so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

Therefore, we need to further restrict the domain to those values in $(-\infty, 1) \cup (1, \infty)$ that make

$f(c) \ge 0$

Equivalently,

$$\frac{1}{c-1}$\ge 0$

Since 1 is positive, and the quotient is nonnegative as well, it follows that

,

and

.

The domain of $g\circ f$ is therefore $(1, \infty)$ .

Define functions $f(x) = $\frac{1}{x-1}$$ and $g(x)= $\sqrt{x}$+ 1$ .

Give the domain of the composition $f \circ g$ .

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The domain of the composition of functions $f \circ g$ is the set of all values that fall in the domain of such that falls in the domain of .

$g(x)= $\sqrt{x}$+ 1$ , a square root function, so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

$f(x) = $\frac{1}{x-1}$$ , a rational function, so its domain is the set of all numbers except those that make its denominator zero. if and only if , so from $\left [ 0, \infty\right )$ , we exclude only the value so that

;

that is,

$$\sqrt{c}$+ 1 = 1$

$$\sqrt{c}$+ 1 - 1 = 1 - 1$

$$\sqrt{c}$ = 0$

Excluding this value from the domain, this leaves $\left ( 0, \infty\right )$ as the domain of $f \circ g$ .

Define functions $f(x) = $x^{2}$-16$ and $g(x) = $\sqrt{x}$$ .

Give the domain of the composition $g\circ f$ .

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The domain of the composition of functions $g\circ f$ is that the set of all values that fall in the domain of such that falls in the domain of .

$f(x) = $x^{2}$-16$ , a polynomial function, so its domain is the set of all real numbers.

$g(x) = $\sqrt{x}$$ , so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

Therefore, the domain of $g\circ f$ is the set of all real numbers such that

$f(c) \ge 0$ .

Equivalently,

To solve this, first, solve the corresponding equation:

Factor the polynomial at left using the difference of squares pattern:

By the Zero Factor Property, either

, in which case ,

or

, in which case .

These are the boundary values of the intervals to be tested. Choose an interior value of each interval and test it in the inequality:

$(-\infty, -4)$

Testing :

$25-16 \ge 0$

$9 \ge 0$ - True; include

Testing :

$0 -16 \ge 0$

$-16 \ge 0$ - False; exclude

$( 4 , \infty)$

Testing :

$25-16 \ge 0$

$9 \ge 0$ - True; include

Include $(-\infty, -4)$ and $( 4 , \infty)$ . Also include the boundary values, since equality is included in the inequality statement ( $\ge$ ).

The correct domain is $(-\infty, -4] \cup [4 , \infty)$ .

Define functions $f(x) = $x^{2}$-16$ and $g(x) = $\sqrt{x}$$ .

Give the domain of the composition $f \circ g$ .

Tap to see back →

The domain of the composition of functions $f \circ g$ is the set of all values that fall in the domain of such that falls in the domain of .

$g(x) = $\sqrt{x}$$ , a square root function, so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

$f(x) = $x^{2}$-16$ , a polynomial function; its domain is the set of all reals, so it does not restrict the domain any further.

The correct domain is $\left [ 0, \infty\right )$ .

Define functions $f(x) = $\sqrt{x-8}$$ and $g(x) = $\sqrt{x+ 8}$$ .

Give the domain of the function .

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The domain of the sum of two functions is the intersections of the domains of the individual functions. Both and are square root functions, so their radicands must both be positive.

The domain of is the set of all values of such that:

$x-8 \ge 0$

Adding 8 to both sides:

$x-8 + 8 \ge 0 + 8$

$x \ge 8$

This domain is $[8, \infty )$ .

The domain of is the set of all values of such that:

$x+8 \ge 0$

Subtracting 8 tfrom both sides:

$x+ 8-8 \ge 0 - 8$

$x \ge -8$

This domain is $[-8, \infty )$

The domain of is the intersection of the domains, which is $[8, \infty )$ .

Define functions $f(x) = $\sqrt{x+8}$$ and $g(x) = $\sqrt{8-x}$$ .

Give the domain of the function .

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The domain of the difference of two functions is the intersections of the domains of the individual functions. Both and are square root functions, so their radicands must both be positive.

The domain of is the set of all values of such that:

$x+8 \ge 0$

Subtracting 8 tfrom both sides:

$x+ 8-8 \ge 0 - 8$

$x \ge -8$

This domain is $[-8, \infty )$

The domain of is the set of all values of such that:

$8-x \ge 0$

Adding to both sides:

$8-x + x \ge 0 + x$

$8 \ge x$ , or

$x \le 8$

This domain is $(-\infty , 8]$ .

The domain of is the intersection of these, .

Define functions $f(x) = \sqrt[3]{x-8}$ and $g(x) = \sqrt[3]{x+ 8}$ .

Give the domain of the function .

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The domain of the sum of two functions is the intersections of the domains of the individual functions.

Both and are cube root functions. There is no restriction on the value of the radicand of a cube root, so both of these functions have as their domain the set of all real numbers $(-\infty, \infty)$ . It follows that also has domain $(-\infty, \infty)$ .

Define functions $f(x) = $\frac{1}{x-1}$$ and $g(x)= $\sqrt{x}$+ 1$ .

Give the domain of the composition $f \circ g$ .

Tap to see back →

The domain of the composition of functions $f \circ g$ is the set of all values that fall in the domain of such that falls in the domain of .

$g(x)= $\sqrt{x}$+ 1$ , a square root function, so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

$f(x) = $\frac{1}{x-1}$$ , a rational function, so its domain is the set of all numbers except those that make its denominator zero. if and only if , so from $\left [ 0, \infty\right )$ , we exclude only the value so that

;

that is,

$$\sqrt{c}$+ 1 = 1$

$$\sqrt{c}$+ 1 - 1 = 1 - 1$

$$\sqrt{c}$ = 0$

Excluding this value from the domain, this leaves $\left ( 0, \infty\right )$ as the domain of $f \circ g$ .

Define functions $f(x) = $\frac{1}{x-1}$$ and $g(x)= $\sqrt{x}$+ 1$ .

Give the domain of the composition $g\circ f$ .

Tap to see back →

The domain of the composition of functions $g\circ f$ is that the set of all values that fall in the domain of such that falls in the domain of .

$f(x) = $\frac{1}{x-1}$$ , a rational function, so its domain is the set of all numbers except those that make its denominator zero. if and only if , making its domain $(-\infty, 1) \cup (1, \infty)$ .

$g(x)= $\sqrt{x}$+ 1$ , a square root function, so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

Therefore, we need to further restrict the domain to those values in $(-\infty, 1) \cup (1, \infty)$ that make

$f(c) \ge 0$

Equivalently,

$$\frac{1}{c-1}$\ge 0$

Since 1 is positive, and the quotient is nonnegative as well, it follows that

,

and

.

The domain of $g\circ f$ is therefore $(1, \infty)$ .

Define functions $f(x) = $\sqrt{x}$$ and $g(x) =- $x^{2}$$

Give the domain of the composition $g\circ f$ .

Tap to see back →

The domain of the composition of functions $g\circ f$ is that the set of all values that fall in the domain of such that falls in the domain of .

$f(x) = $\sqrt{x}$$ , a square root function, whose domain is the set of all values of that make the radicand nonnegative. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

$g(x) =- $x^{2}$$ , a polynomial function, whose domain is the set of all real numbers. Therefore, the domain is not restricted further by . The domain of $g\circ f$ is $\left [ 0, \infty\right )$ .

Define functions $f(x) = $x^{2}$-16$ and $g(x) = $\sqrt{x}$$ .

Give the domain of the composition $g\circ f$ .

Tap to see back →

The domain of the composition of functions $g\circ f$ is that the set of all values that fall in the domain of such that falls in the domain of .

$f(x) = $x^{2}$-16$ , a polynomial function, so its domain is the set of all real numbers.

$g(x) = $\sqrt{x}$$ , so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

Therefore, the domain of $g\circ f$ is the set of all real numbers such that

$f(c) \ge 0$ .

Equivalently,

To solve this, first, solve the corresponding equation:

Factor the polynomial at left using the difference of squares pattern:

By the Zero Factor Property, either

, in which case ,

or

, in which case .

These are the boundary values of the intervals to be tested. Choose an interior value of each interval and test it in the inequality:

$(-\infty, -4)$

Testing :

$25-16 \ge 0$

$9 \ge 0$ - True; include

Testing :

$0 -16 \ge 0$

$-16 \ge 0$ - False; exclude

$( 4 , \infty)$

Testing :

$25-16 \ge 0$

$9 \ge 0$ - True; include

Include $(-\infty, -4)$ and $( 4 , \infty)$ . Also include the boundary values, since equality is included in the inequality statement ( $\ge$ ).

The correct domain is $(-\infty, -4] \cup [4 , \infty)$ .

Define functions $f(x) = $x^{2}$-16$ and $g(x) = $\sqrt{x}$$ .

Give the domain of the composition $f \circ g$ .

Tap to see back →

The domain of the composition of functions $f \circ g$ is the set of all values that fall in the domain of such that falls in the domain of .

$g(x) = $\sqrt{x}$$ , a square root function, so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

$f(x) = $x^{2}$-16$ , a polynomial function; its domain is the set of all reals, so it does not restrict the domain any further.

The correct domain is $\left [ 0, \infty\right )$ .

Define functions $f(x) = $\sqrt{x-8}$$ and $g(x) = $\sqrt{x+ 8}$$ .

Give the domain of the function .

Tap to see back →

The domain of the sum of two functions is the intersections of the domains of the individual functions. Both and are square root functions, so their radicands must both be positive.

The domain of is the set of all values of such that:

$x-8 \ge 0$

Adding 8 to both sides:

$x-8 + 8 \ge 0 + 8$

$x \ge 8$

This domain is $[8, \infty )$ .

The domain of is the set of all values of such that:

$x+8 \ge 0$

Subtracting 8 tfrom both sides:

$x+ 8-8 \ge 0 - 8$

$x \ge -8$

This domain is $[-8, \infty )$

The domain of is the intersection of the domains, which is $[8, \infty )$ .

Define functions $f(x) = $\sqrt{x+8}$$ and $g(x) = $\sqrt{8-x}$$ .

Give the domain of the function .

Tap to see back →

The domain of the difference of two functions is the intersections of the domains of the individual functions. Both and are square root functions, so their radicands must both be positive.

The domain of is the set of all values of such that:

$x+8 \ge 0$

Subtracting 8 tfrom both sides:

$x+ 8-8 \ge 0 - 8$

$x \ge -8$

This domain is $[-8, \infty )$

The domain of is the set of all values of such that:

$8-x \ge 0$

Adding to both sides:

$8-x + x \ge 0 + x$

$8 \ge x$ , or

$x \le 8$

This domain is $(-\infty , 8]$ .

The domain of is the intersection of these, .

Define functions $f(x) = \sqrt[3]{x-8}$ and $g(x) = \sqrt[3]{x+ 8}$ .

Give the domain of the function .

Tap to see back →

The domain of the sum of two functions is the intersections of the domains of the individual functions.

Both and are cube root functions. There is no restriction on the value of the radicand of a cube root, so both of these functions have as their domain the set of all real numbers $(-\infty, \infty)$ . It follows that also has domain $(-\infty, \infty)$ .

Define functions $f(x) = $\sqrt{x}$$ and $g(x) =- $x^{2}$$

Give the domain of the composition $g\circ f$ .

Tap to see back →

The domain of the composition of functions $g\circ f$ is that the set of all values that fall in the domain of such that falls in the domain of .

$f(x) = $\sqrt{x}$$ , a square root function, whose domain is the set of all values of that make the radicand nonnegative. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

$g(x) =- $x^{2}$$ , a polynomial function, whose domain is the set of all real numbers. Therefore, the domain is not restricted further by . The domain of $g\circ f$ is $\left [ 0, \infty\right )$ .

Define functions $f(x) = $\frac{1}{x-1}$$ and $g(x)= $\sqrt{x}$+ 1$ .

Give the domain of the composition $g\circ f$ .

Tap to see back →

The domain of the composition of functions $g\circ f$ is that the set of all values that fall in the domain of such that falls in the domain of .

$f(x) = $\frac{1}{x-1}$$ , a rational function, so its domain is the set of all numbers except those that make its denominator zero. if and only if , making its domain $(-\infty, 1) \cup (1, \infty)$ .

$g(x)= $\sqrt{x}$+ 1$ , a square root function, so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

Therefore, we need to further restrict the domain to those values in $(-\infty, 1) \cup (1, \infty)$ that make

$f(c) \ge 0$

Equivalently,

$$\frac{1}{c-1}$\ge 0$

Since 1 is positive, and the quotient is nonnegative as well, it follows that

,

and

.

The domain of $g\circ f$ is therefore $(1, \infty)$ .

Define functions $f(x) = $\frac{1}{x-1}$$ and $g(x)= $\sqrt{x}$+ 1$ .

Give the domain of the composition $f \circ g$ .

Tap to see back →

The domain of the composition of functions $f \circ g$ is the set of all values that fall in the domain of such that falls in the domain of .

$g(x)= $\sqrt{x}$+ 1$ , a square root function, so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

$f(x) = $\frac{1}{x-1}$$ , a rational function, so its domain is the set of all numbers except those that make its denominator zero. if and only if , so from $\left [ 0, \infty\right )$ , we exclude only the value so that

;

that is,

$$\sqrt{c}$+ 1 = 1$

$$\sqrt{c}$+ 1 - 1 = 1 - 1$

$$\sqrt{c}$ = 0$

Excluding this value from the domain, this leaves $\left ( 0, \infty\right )$ as the domain of $f \circ g$ .

Define functions $f(x) = $x^{2}$-16$ and $g(x) = $\sqrt{x}$$ .

Give the domain of the composition $g\circ f$ .

Tap to see back →

The domain of the composition of functions $g\circ f$ is that the set of all values that fall in the domain of such that falls in the domain of .

$f(x) = $x^{2}$-16$ , a polynomial function, so its domain is the set of all real numbers.

$g(x) = $\sqrt{x}$$ , so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

Therefore, the domain of $g\circ f$ is the set of all real numbers such that

$f(c) \ge 0$ .

Equivalently,

To solve this, first, solve the corresponding equation:

Factor the polynomial at left using the difference of squares pattern:

By the Zero Factor Property, either

, in which case ,

or

, in which case .

These are the boundary values of the intervals to be tested. Choose an interior value of each interval and test it in the inequality:

$(-\infty, -4)$

Testing :

$25-16 \ge 0$

$9 \ge 0$ - True; include

Testing :

$0 -16 \ge 0$

$-16 \ge 0$ - False; exclude

$( 4 , \infty)$

Testing :

$25-16 \ge 0$

$9 \ge 0$ - True; include

Include $(-\infty, -4)$ and $( 4 , \infty)$ . Also include the boundary values, since equality is included in the inequality statement ( $\ge$ ).

The correct domain is $(-\infty, -4] \cup [4 , \infty)$ .