QuestionDetermine the domain and range of the function $h(x)=\frac{1}{\sqrt[4]{x^{2}-5 x}}$ .

Studdy Solution

STEP 1

Assumptions1. The function is $h(x)=\frac{1}{\sqrt[4]{x^{}-5 x}}$ . We need to find the domain and range of this function

STEP 2

The domain of a function is the set of all possible input values (typically the "x" variable), which produce a valid output from a particular function.For the given function, the denominator $\sqrt[4]{x^{2}-5 x}$ cannot be zero or negative, as we cannot divide by zero and we cannot take the fourth root of a negative number.
So, we need to solve the inequality $x^{2}-5 x >0$ to find the domain.

STEP 3

To solve the inequality $x^{2}-5 x >0$ , we first factor the left side.
$x(x-5) >0$

STEP 4

Now we need to find the critical points, which are the solutions to $x(x-) =0$ .
Setting each factor equal to zero gives us $x=0$ and $x=$ .

STEP 5

We then test the intervals defined by these critical points in the inequality $x(x-5) >0$ .
For $x<0$ , both factors are negative, so the product is positive.
For $0<x<5$ , the first factor is positive and the second is negative, so the product is negative.
For $x>5$ , both factors are positive, so the product is positive.

STEP 6

Therefore, the solution to the inequality $x(x-5) >0$ is $x<0$ or $x>5$ . This is the domain of the function.

STEP 7

The range of a function is the set of all possible output values (typically the "y" variable), which are the result of the function.
For the given function, since the denominator is always positive (from the domain), the function is always positive.

STEP 8

Also, as $x$ approaches the endpoints of the domain (0 and5 from the left and right, respectively), the function approaches infinity.

STEP 9

Therefore, the range of the function is $(, \infty)$ .
The domain of the function $h(x)=\frac{}{\sqrt[4]{x^{2}-5 x}}$ is $x<$ or $x>5$ , and the range is $(, \infty)$ .

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QuestionDetermine the domain and range of the function $h(x)=\frac{1}{\sqrt[4]{x^{2}-5 x}}$ .

Studdy Solution

STEP 1

Assumptions1. The function is $h(x)=\frac{1}{\sqrt[4]{x^{}-5 x}}$ . We need to find the domain and range of this function

STEP 2

STEP 3

To solve the inequality $x^{2}-5 x >0$ , we first factor the left side.
$x(x-5) >0$

STEP 4

Now we need to find the critical points, which are the solutions to $x(x-) =0$ .
Setting each factor equal to zero gives us $x=0$ and $x=$ .

STEP 5

STEP 6

Therefore, the solution to the inequality $x(x-5) >0$ is $x<0$ or $x>5$ . This is the domain of the function.

STEP 7

The range of a function is the set of all possible output values (typically the "y" variable), which are the result of the function.
For the given function, since the denominator is always positive (from the domain), the function is always positive.

STEP 8

Also, as $x$ approaches the endpoints of the domain (0 and5 from the left and right, respectively), the function approaches infinity.

STEP 9

Therefore, the range of the function is $(, \infty)$ .
The domain of the function $h(x)=\frac{}{\sqrt[4]{x^{2}-5 x}}$ is $x<$ or $x>5$ , and the range is $(, \infty)$ .

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QuestionDetermine the domain and range of the function h(x)=1x2−5x4h(x)=\frac{1}{\sqrt[4]{x^{2}-5 x}}h(x)=4x2−5x​1​.

Studdy Solution

STEP 1

Assumptions1. The function is h(x)=1x−5x4h(x)=\frac{1}{\sqrt[4]{x^{}-5 x}}h(x)=4x−5x​1​ . We need to find the domain and range of this function

STEP 2

STEP 3

To solve the inequality x2−5x>0x^{2}-5 x >0x2−5x>0, we first factor the left side.x(x−5)>0x(x-5) >0x(x−5)>0

STEP 4

Now we need to find the critical points, which are the solutions to x(x−)=0x(x-) =0x(x−)=0.Setting each factor equal to zero gives us x=0x=0x=0 and x=x=x=.

STEP 5

STEP 6

Therefore, the solution to the inequality x(x−5)>0x(x-5) >0x(x−5)>0 is x<0x<0x<0 or x>5x>5x>5. This is the domain of the function.

STEP 7

The range of a function is the set of all possible output values (typically the "y" variable), which are the result of the function.For the given function, since the denominator is always positive (from the domain), the function is always positive.

STEP 8

Also, as xxx approaches the endpoints of the domain (0 and5 from the left and right, respectively), the function approaches infinity.

STEP 9

Therefore, the range of the function is (,∞)(, \infty)(,∞).The domain of the function h(x)=x2−5x4h(x)=\frac{}{\sqrt[4]{x^{2}-5 x}}h(x)=4x2−5x​​ is x<x<x< or x>5x>5x>5, and the range is (,∞)(, \infty)(,∞).

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QuestionDetermine the domain and range of the function $h(x)=\frac{1}{\sqrt[4]{x^{2}-5 x}}$ .

Assumptions1. The function is $h(x)=\frac{1}{\sqrt[4]{x^{}-5 x}}$ . We need to find the domain and range of this function

To solve the inequality $x^{2}-5 x >0$ , we first factor the left side.
$x(x-5) >0$

Now we need to find the critical points, which are the solutions to $x(x-) =0$ .
Setting each factor equal to zero gives us $x=0$ and $x=$ .

Therefore, the solution to the inequality $x(x-5) >0$ is $x<0$ or $x>5$ . This is the domain of the function.

The range of a function is the set of all possible output values (typically the "y" variable), which are the result of the function.
For the given function, since the denominator is always positive (from the domain), the function is always positive.

Also, as $x$ approaches the endpoints of the domain (0 and5 from the left and right, respectively), the function approaches infinity.

Therefore, the range of the function is $(, \infty)$ .
The domain of the function $h(x)=\frac{}{\sqrt[4]{x^{2}-5 x}}$ is $x<$ or $x>5$ , and the range is $(, \infty)$ .