QuestionFind the domain of each function: 48. $f(x)=x^{2}+2$ , 52. $h(x)=\frac{2x}{x^{2}-4}$ , 54. $G(x)=\frac{x+4}{x^{3}-4x}$ , 56. $G(x)=\sqrt{1-x}$ , 58. $f(x)=\frac{x}{\sqrt{x-4}}$ .

Studdy Solution

STEP 1

Assumptions1. We are finding the domain of each function, which is the set of all possible input values (x-values) that will output real numbers. . The functions are real-valued functions.

STEP 2

For the function $f(x)=x^{2}+2$ , the domain is all real numbers because for any real number x, $x^{2}+2$ will be a real number.
$Domain\, of\, f(x)=(-\infty, \infty)$

STEP 3

For the function $h(x)=\frac{2 x}{x^{2}-}$ , the denominator cannot be zero because division by zero is undefined. So, we need to find the values of x that make the denominator zero and exclude them from the domain.
$x^{2}-=0$

STEP 4

olve the equation $x^{2}-4=0$ for x.
$x^{2}=4$ $x=\pm\sqrt{4}$ $x=\pm2$

STEP 5

The values $x=2$ and $x=-2$ make the denominator zero, so they are not in the domain. The domain of $h(x)$ is all real numbers except $x=2$ and $x=-2$ .
$Domain\, of\, h(x)=(-\infty, -2) \cup (-2,2) \cup (2, \infty)$

STEP 6

For the function $G(x)=\frac{x+4}{x^{3}-4 x}$ , the denominator cannot be zero. So, we need to find the values of x that make the denominator zero and exclude them from the domain.
$x^{3}-4x=0$

STEP 7

Factor the equation $x^{3}-4x=0$ to find the roots.
$x(x^{2}-4)=0$ $x(x-\sqrt{4})(x+\sqrt{4})=0$ $x(x-2)(x+2)=0$

STEP 8

olve the equation $x(x-2)(x+2)=0$ for x.
$x=0, x=2, x=-2$

STEP 9

The values $x=$ , $x=2$ , and $x=-2$ make the denominator zero, so they are not in the domain. The domain of $G(x)$ is all real numbers except $x=$ , $x=2$ , and $x=-2$ .
$Domain\, of\, G(x)=(-\infty, -2) \cup (-2,) \cup (,2) \cup (2, \infty)$

STEP 10

For the function $G(x)=\sqrt{-x}$ , the expression under the square root cannot be negative because the square root of a negative number is not a real number. So, we need to find the values of x that make the expression under the square root nonnegative.
$-x \geq0$

STEP 11

olve the inequality $-x \geq0$ for x.
$x \leq$

STEP 12

The domain of $G(x)$ is all real numbers less than or equal to.
$Domain\, of\, G(x)=(-\infty,]$

STEP 13

For the function $f(x)=\frac{x}{\sqrt{x-}}$ , the expression under the square root cannot be negative and the denominator cannot be zero. So, we need to find the values of x that make the expression under the square root positive.
$x- >0$

STEP 14

olve the inequality $x-4 >0$ for x.
$x >4$

STEP 15

The domain of $f(x)$ is all real numbers greater than4.
$Domain\, of\, f(x)=(4, \infty)$

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QuestionFind the domain of each function: 48. $f(x)=x^{2}+2$ , 52. $h(x)=\frac{2x}{x^{2}-4}$ , 54. $G(x)=\frac{x+4}{x^{3}-4x}$ , 56. $G(x)=\sqrt{1-x}$ , 58. $f(x)=\frac{x}{\sqrt{x-4}}$ .

Studdy Solution

STEP 1

Assumptions1. We are finding the domain of each function, which is the set of all possible input values (x-values) that will output real numbers. . The functions are real-valued functions.

STEP 2

For the function $f(x)=x^{2}+2$ , the domain is all real numbers because for any real number x, $x^{2}+2$ will be a real number.
$Domain\, of\, f(x)=(-\infty, \infty)$

STEP 3

For the function $h(x)=\frac{2 x}{x^{2}-}$ , the denominator cannot be zero because division by zero is undefined. So, we need to find the values of x that make the denominator zero and exclude them from the domain.
$x^{2}-=0$

STEP 4

olve the equation $x^{2}-4=0$ for x.
$x^{2}=4$ $x=\pm\sqrt{4}$ $x=\pm2$

STEP 5

The values $x=2$ and $x=-2$ make the denominator zero, so they are not in the domain. The domain of $h(x)$ is all real numbers except $x=2$ and $x=-2$ .
$Domain\, of\, h(x)=(-\infty, -2) \cup (-2,2) \cup (2, \infty)$

STEP 6

For the function $G(x)=\frac{x+4}{x^{3}-4 x}$ , the denominator cannot be zero. So, we need to find the values of x that make the denominator zero and exclude them from the domain.
$x^{3}-4x=0$

STEP 7

Factor the equation $x^{3}-4x=0$ to find the roots.
$x(x^{2}-4)=0$ $x(x-\sqrt{4})(x+\sqrt{4})=0$ $x(x-2)(x+2)=0$

STEP 8

olve the equation $x(x-2)(x+2)=0$ for x.
$x=0, x=2, x=-2$

STEP 9

The values $x=$ , $x=2$ , and $x=-2$ make the denominator zero, so they are not in the domain. The domain of $G(x)$ is all real numbers except $x=$ , $x=2$ , and $x=-2$ .
$Domain\, of\, G(x)=(-\infty, -2) \cup (-2,) \cup (,2) \cup (2, \infty)$

STEP 10

For the function $G(x)=\sqrt{-x}$ , the expression under the square root cannot be negative because the square root of a negative number is not a real number. So, we need to find the values of x that make the expression under the square root nonnegative.
$-x \geq0$

STEP 11

olve the inequality $-x \geq0$ for x.
$x \leq$

STEP 12

The domain of $G(x)$ is all real numbers less than or equal to.
$Domain\, of\, G(x)=(-\infty,]$

STEP 13

For the function $f(x)=\frac{x}{\sqrt{x-}}$ , the expression under the square root cannot be negative and the denominator cannot be zero. So, we need to find the values of x that make the expression under the square root positive.
$x- >0$

STEP 14

olve the inequality $x-4 >0$ for x.
$x >4$

STEP 15

The domain of $f(x)$ is all real numbers greater than4.
$Domain\, of\, f(x)=(4, \infty)$

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Studdy Solution

STEP 1

Assumptions1. We are finding the domain of each function, which is the set of all possible input values (x-values) that will output real numbers. . The functions are real-valued functions.

STEP 2

For the function f(x)=x2+2f(x)=x^{2}+2f(x)=x2+2, the domain is all real numbers because for any real number x, x2+2x^{2}+2x2+2 will be a real number.Domain of f(x)=(−∞,∞)Domain\, of\, f(x)=(-\infty, \infty)Domainoff(x)=(−∞,∞)

STEP 3

For the function h(x)=2xx2−h(x)=\frac{2 x}{x^{2}-}h(x)=x2−2x​, the denominator cannot be zero because division by zero is undefined. So, we need to find the values of x that make the denominator zero and exclude them from the domain.x2−=0x^{2}-=0x2−=0

STEP 4

olve the equation x2−4=0x^{2}-4=0x2−4=0 for x.x2=4x^{2}=4x2=4x=±4x=\pm\sqrt{4}x=±4​x=±2x=\pm2x=±2

STEP 5

STEP 6

For the function G(x)=x+4x3−4xG(x)=\frac{x+4}{x^{3}-4 x}G(x)=x3−4xx+4​, the denominator cannot be zero. So, we need to find the values of x that make the denominator zero and exclude them from the domain.x3−4x=0x^{3}-4x=0x3−4x=0

STEP 7

Factor the equation x3−4x=0x^{3}-4x=0x3−4x=0 to find the roots.x(x2−4)=0x(x^{2}-4)=0x(x2−4)=0x(x−4)(x+4)=0x(x-\sqrt{4})(x+\sqrt{4})=0x(x−4​)(x+4​)=0x(x−2)(x+2)=0x(x-2)(x+2)=0x(x−2)(x+2)=0

STEP 8

olve the equation x(x−2)(x+2)=0x(x-2)(x+2)=0x(x−2)(x+2)=0 for x.x=0,x=2,x=−2x=0, x=2, x=-2x=0,x=2,x=−2

STEP 9

STEP 10

For the function G(x)=−xG(x)=\sqrt{-x}G(x)=−x​, the expression under the square root cannot be negative because the square root of a negative number is not a real number. So, we need to find the values of x that make the expression under the square root nonnegative.−x≥0-x \geq0−x≥0

STEP 11

olve the inequality −x≥0-x \geq0−x≥0 for x.x≤x \leqx≤

STEP 12

The domain of G(x)G(x)G(x) is all real numbers less than or equal to.Domain of G(x)=(−∞,]Domain\, of\, G(x)=(-\infty,]DomainofG(x)=(−∞,]

STEP 13

For the function f(x)=xx−f(x)=\frac{x}{\sqrt{x-}}f(x)=x−​x​, the expression under the square root cannot be negative and the denominator cannot be zero. So, we need to find the values of x that make the expression under the square root positive.x−>0x- >0x−>0

STEP 14

olve the inequality x−4>0x-4 >0x−4>0 for x.x>4x >4x>4

STEP 15

The domain of f(x)f(x)f(x) is all real numbers greater than4.Domain of f(x)=(4,∞)Domain\, of\, f(x)=(4, \infty)Domainoff(x)=(4,∞)

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For the function $f(x)=x^{2}+2$ , the domain is all real numbers because for any real number x, $x^{2}+2$ will be a real number.
$Domain\, of\, f(x)=(-\infty, \infty)$

For the function $h(x)=\frac{2 x}{x^{2}-}$ , the denominator cannot be zero because division by zero is undefined. So, we need to find the values of x that make the denominator zero and exclude them from the domain.
$x^{2}-=0$

olve the equation $x^{2}-4=0$ for x.
$x^{2}=4$ $x=\pm\sqrt{4}$ $x=\pm2$

For the function $G(x)=\frac{x+4}{x^{3}-4 x}$ , the denominator cannot be zero. So, we need to find the values of x that make the denominator zero and exclude them from the domain.
$x^{3}-4x=0$

Factor the equation $x^{3}-4x=0$ to find the roots.
$x(x^{2}-4)=0$ $x(x-\sqrt{4})(x+\sqrt{4})=0$ $x(x-2)(x+2)=0$

olve the equation $x(x-2)(x+2)=0$ for x.
$x=0, x=2, x=-2$

For the function $G(x)=\sqrt{-x}$ , the expression under the square root cannot be negative because the square root of a negative number is not a real number. So, we need to find the values of x that make the expression under the square root nonnegative.
$-x \geq0$

olve the inequality $-x \geq0$ for x.
$x \leq$

The domain of $G(x)$ is all real numbers less than or equal to.
$Domain\, of\, G(x)=(-\infty,]$

For the function $f(x)=\frac{x}{\sqrt{x-}}$ , the expression under the square root cannot be negative and the denominator cannot be zero. So, we need to find the values of x that make the expression under the square root positive.
$x- >0$

olve the inequality $x-4 >0$ for x.
$x >4$

The domain of $f(x)$ is all real numbers greater than4.
$Domain\, of\, f(x)=(4, \infty)$