IDNLearn.com connects you with experts who provide accurate and reliable answers. Find the answers you need quickly and accurately with help from our knowledgeable and dedicated community members.
Sagot :
To identify the domain and range of the function [tex]\( y = \sqrt[3]{\frac{x-1}{8}} + 1 \)[/tex] and then sketch its graph, we will follow these steps:
### Identifying the Domain
The domain of a function is the set of all possible input values (x-values) for which the function is defined. In this specific case, the function involves the cube root. The cube root function, [tex]\(\sqrt[3]{z}\)[/tex], is defined for all real numbers [tex]\(z\)[/tex].
The argument inside the cube root here is [tex]\(\frac{x-1}{8}\)[/tex]. Since the cube root is defined for all real numbers, there are no restrictions on [tex]\(x\)[/tex]. This means that [tex]\(x\)[/tex] can be any real number. Therefore, the domain of the function [tex]\( y = \sqrt[3]{\frac{x-1}{8}} + 1 \)[/tex] is all real numbers.
Domain: all real numbers ([tex]\( -\infty < x < \infty \)[/tex])
### Identifying the Range
The range of a function is the set of all possible output values (y-values) the function can take. For the cube root function [tex]\(\sqrt[3]{z}\)[/tex], as [tex]\(z\)[/tex] varies over all real numbers, the cube root can also take on all real number values.
Given that the function [tex]\( y = \sqrt[3]{\frac{x-1}{8}} + 1 \)[/tex] involves adding 1 to the cube root, the output value is also capable of taking on any real number.
Range: all real numbers ([tex]\( -\infty < y < \infty \)[/tex])
### Sketching the Graph
To sketch the graph, follow these steps:
1. Identify the basic shape of the cube root function: The basic cube root function [tex]\( y = \sqrt[3]{x} \)[/tex] has an S-shape. Some key points on this graph are:
- [tex]\( (-1, -1) \)[/tex]
- [tex]\( (0, 0) \)[/tex]
- [tex]\( (1, 1) \)[/tex]
2. Apply the transformations:
- Translation transformation: The function [tex]\( y = \sqrt[3]{\frac{x-1}{8}} + 1 \)[/tex] can be seen as a transformation of [tex]\( y = \sqrt[3]{x} \)[/tex]:
- The term [tex]\(\frac{x-1}{8}\)[/tex] shifts the graph horizontally.
- [tex]\( \frac{x-1}{8} = 0 \)[/tex] when [tex]\( x = 1 \)[/tex]. So, there is a horizontal shift of 1 unit to the right.
- The addition of 1 to the cube root function shifts the graph vertically.
- Adding 1 to the function results in a vertical shift of 1 unit up.
3. Plot the transformed points:
- Start with the reference points of the base graph and then apply the transformations to them:
- [tex]\( (-1, -1) \)[/tex] transforms to [tex]\( (-1+1, -1+1) = (0, 0) \)[/tex]
- [tex]\( (0, 0) \)[/tex] transforms to [tex]\( (0+1, 0+1) = (1, 1) \)[/tex]
- [tex]\( (1, 1) \)[/tex] transforms to [tex]\( (1+1, 1+1) = (2, 2) \)[/tex]
4. Draw the curve:
- Sketch the S-shaped curve passing through the points [tex]\( (0, 0) \)[/tex], [tex]\( (1, 1) \)[/tex], and [tex]\( (2, 2) \)[/tex]. Ensure it follows the nature of the cube root function (flattening out as it moves away from the origin).
By following these steps, you should have a comprehensive understanding and a sketch of the function [tex]\( y = \sqrt[3]{\frac{x-1}{8}} + 1 \)[/tex].
### Identifying the Domain
The domain of a function is the set of all possible input values (x-values) for which the function is defined. In this specific case, the function involves the cube root. The cube root function, [tex]\(\sqrt[3]{z}\)[/tex], is defined for all real numbers [tex]\(z\)[/tex].
The argument inside the cube root here is [tex]\(\frac{x-1}{8}\)[/tex]. Since the cube root is defined for all real numbers, there are no restrictions on [tex]\(x\)[/tex]. This means that [tex]\(x\)[/tex] can be any real number. Therefore, the domain of the function [tex]\( y = \sqrt[3]{\frac{x-1}{8}} + 1 \)[/tex] is all real numbers.
Domain: all real numbers ([tex]\( -\infty < x < \infty \)[/tex])
### Identifying the Range
The range of a function is the set of all possible output values (y-values) the function can take. For the cube root function [tex]\(\sqrt[3]{z}\)[/tex], as [tex]\(z\)[/tex] varies over all real numbers, the cube root can also take on all real number values.
Given that the function [tex]\( y = \sqrt[3]{\frac{x-1}{8}} + 1 \)[/tex] involves adding 1 to the cube root, the output value is also capable of taking on any real number.
Range: all real numbers ([tex]\( -\infty < y < \infty \)[/tex])
### Sketching the Graph
To sketch the graph, follow these steps:
1. Identify the basic shape of the cube root function: The basic cube root function [tex]\( y = \sqrt[3]{x} \)[/tex] has an S-shape. Some key points on this graph are:
- [tex]\( (-1, -1) \)[/tex]
- [tex]\( (0, 0) \)[/tex]
- [tex]\( (1, 1) \)[/tex]
2. Apply the transformations:
- Translation transformation: The function [tex]\( y = \sqrt[3]{\frac{x-1}{8}} + 1 \)[/tex] can be seen as a transformation of [tex]\( y = \sqrt[3]{x} \)[/tex]:
- The term [tex]\(\frac{x-1}{8}\)[/tex] shifts the graph horizontally.
- [tex]\( \frac{x-1}{8} = 0 \)[/tex] when [tex]\( x = 1 \)[/tex]. So, there is a horizontal shift of 1 unit to the right.
- The addition of 1 to the cube root function shifts the graph vertically.
- Adding 1 to the function results in a vertical shift of 1 unit up.
3. Plot the transformed points:
- Start with the reference points of the base graph and then apply the transformations to them:
- [tex]\( (-1, -1) \)[/tex] transforms to [tex]\( (-1+1, -1+1) = (0, 0) \)[/tex]
- [tex]\( (0, 0) \)[/tex] transforms to [tex]\( (0+1, 0+1) = (1, 1) \)[/tex]
- [tex]\( (1, 1) \)[/tex] transforms to [tex]\( (1+1, 1+1) = (2, 2) \)[/tex]
4. Draw the curve:
- Sketch the S-shaped curve passing through the points [tex]\( (0, 0) \)[/tex], [tex]\( (1, 1) \)[/tex], and [tex]\( (2, 2) \)[/tex]. Ensure it follows the nature of the cube root function (flattening out as it moves away from the origin).
By following these steps, you should have a comprehensive understanding and a sketch of the function [tex]\( y = \sqrt[3]{\frac{x-1}{8}} + 1 \)[/tex].
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com is your go-to source for dependable answers. Thank you for visiting, and we hope to assist you again.
