Get the most out of your questions with IDNLearn.com's extensive resources. Ask any question and receive accurate, in-depth responses from our dedicated team of experts.
Sagot :
To determine whether the function [tex]\( f(x) = x^3 - 1 \)[/tex] is one-to-one, we need to analyze its behavior and check if each output value [tex]\( y \)[/tex] has exactly one corresponding input value [tex]\( x \)[/tex].
### Step 1: Determine if the function is one-to-one
A function is one-to-one if and only if it passes both the horizontal line test and the vertical line test.
#### Vertical Line Test:
Any vertical line drawn through the graph of [tex]\( f(x) \)[/tex] should touch the graph at exactly one point. This is true for [tex]\( f(x) = x^3 - 1 \)[/tex] because it is a polynomial function of degree 3, which does not repeat any values for its input.
#### Horizontal Line Test:
Any horizontal line drawn through the graph of [tex]\( f(x) \)[/tex] should touch the graph at exactly one point. For [tex]\( f(x) = x^3 - 1 \)[/tex], since it is a monotonically increasing function (its derivative [tex]\( f'(x) = 3x^2 \)[/tex] is always non-negative and only equals zero at [tex]\( x = 0 \)[/tex]), it means that each value of [tex]\( f(x) \)[/tex] is achieved exactly once. Therefore, [tex]\( f(x) \)[/tex] passes the horizontal line test as well.
Since [tex]\( f(x) \)[/tex] passes both the vertical and horizontal line tests, the function is one-to-one.
### Step 2: Find the inverse function
To find the inverse function of [tex]\( f(x) = x^3 - 1 \)[/tex], follow these steps:
1. Replace [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = x^3 - 1 \][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y + 1 = x^3 \][/tex]
[tex]\[ x = (y + 1)^{1/3} \][/tex]
3. Replace [tex]\( y \)[/tex] with [tex]\( x \)[/tex] to get the inverse function:
[tex]\[ f^{-1}(x) = (x + 1)^{1/3} \][/tex]
Therefore, the equation for the inverse function is:
[tex]\[ y = f^{-1}(x) = (x + 1)^{1/3} \][/tex]
### Step 3: Graph [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex]
The graph of [tex]\( f(x) = x^3 - 1 \)[/tex] is a cubic curve shifted one unit down. The graph of [tex]\( f^{-1}(x) = (x + 1)^{1/3} \)[/tex] is the cube root function shifted one unit left. When graphed on the same axes, these functions will be symmetric with respect to the line [tex]\( y = x \)[/tex].
### Step 4: Determine the domain and range
- The domain of [tex]\( f(x) = x^3 - 1 \)[/tex] is all real numbers because you can cube any real number.
- The range of [tex]\( f(x) = x^3 - 1 \)[/tex] is all real numbers because the cubic function can produce any real number output.
- The domain of [tex]\( f^{-1}(x) = (x + 1)^{1/3} \)[/tex] is all real numbers because you can take the cube root of any real number.
- The range of [tex]\( f^{-1}(x) = (x + 1)^{1/3} \)[/tex] is all real numbers because the cube root function can produce any real number output.
### Summary
(a) The function [tex]\( f(x) \)[/tex] is one-to-one, and the inverse function is given by:
[tex]\[ f^{-1}(x) = (x + 1)^{1/3} \][/tex]
(b) The graph of [tex]\( f(x) = x^3 - 1 \)[/tex] and [tex]\( f^{-1}(x) = (x + 1)^{1/3} \)[/tex] should be symmetric with respect to the line [tex]\( y = x \)[/tex].
(c) The domain and range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] are:
- Domain of [tex]\( f(x) \)[/tex]: All real numbers
- Range of [tex]\( f(x) \)[/tex]: All real numbers
- Domain of [tex]\( f^{-1}(x) \)[/tex]: All real numbers
- Range of [tex]\( f^{-1}(x) \)[/tex]: All real numbers
### Step 1: Determine if the function is one-to-one
A function is one-to-one if and only if it passes both the horizontal line test and the vertical line test.
#### Vertical Line Test:
Any vertical line drawn through the graph of [tex]\( f(x) \)[/tex] should touch the graph at exactly one point. This is true for [tex]\( f(x) = x^3 - 1 \)[/tex] because it is a polynomial function of degree 3, which does not repeat any values for its input.
#### Horizontal Line Test:
Any horizontal line drawn through the graph of [tex]\( f(x) \)[/tex] should touch the graph at exactly one point. For [tex]\( f(x) = x^3 - 1 \)[/tex], since it is a monotonically increasing function (its derivative [tex]\( f'(x) = 3x^2 \)[/tex] is always non-negative and only equals zero at [tex]\( x = 0 \)[/tex]), it means that each value of [tex]\( f(x) \)[/tex] is achieved exactly once. Therefore, [tex]\( f(x) \)[/tex] passes the horizontal line test as well.
Since [tex]\( f(x) \)[/tex] passes both the vertical and horizontal line tests, the function is one-to-one.
### Step 2: Find the inverse function
To find the inverse function of [tex]\( f(x) = x^3 - 1 \)[/tex], follow these steps:
1. Replace [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = x^3 - 1 \][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y + 1 = x^3 \][/tex]
[tex]\[ x = (y + 1)^{1/3} \][/tex]
3. Replace [tex]\( y \)[/tex] with [tex]\( x \)[/tex] to get the inverse function:
[tex]\[ f^{-1}(x) = (x + 1)^{1/3} \][/tex]
Therefore, the equation for the inverse function is:
[tex]\[ y = f^{-1}(x) = (x + 1)^{1/3} \][/tex]
### Step 3: Graph [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex]
The graph of [tex]\( f(x) = x^3 - 1 \)[/tex] is a cubic curve shifted one unit down. The graph of [tex]\( f^{-1}(x) = (x + 1)^{1/3} \)[/tex] is the cube root function shifted one unit left. When graphed on the same axes, these functions will be symmetric with respect to the line [tex]\( y = x \)[/tex].
### Step 4: Determine the domain and range
- The domain of [tex]\( f(x) = x^3 - 1 \)[/tex] is all real numbers because you can cube any real number.
- The range of [tex]\( f(x) = x^3 - 1 \)[/tex] is all real numbers because the cubic function can produce any real number output.
- The domain of [tex]\( f^{-1}(x) = (x + 1)^{1/3} \)[/tex] is all real numbers because you can take the cube root of any real number.
- The range of [tex]\( f^{-1}(x) = (x + 1)^{1/3} \)[/tex] is all real numbers because the cube root function can produce any real number output.
### Summary
(a) The function [tex]\( f(x) \)[/tex] is one-to-one, and the inverse function is given by:
[tex]\[ f^{-1}(x) = (x + 1)^{1/3} \][/tex]
(b) The graph of [tex]\( f(x) = x^3 - 1 \)[/tex] and [tex]\( f^{-1}(x) = (x + 1)^{1/3} \)[/tex] should be symmetric with respect to the line [tex]\( y = x \)[/tex].
(c) The domain and range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] are:
- Domain of [tex]\( f(x) \)[/tex]: All real numbers
- Range of [tex]\( f(x) \)[/tex]: All real numbers
- Domain of [tex]\( f^{-1}(x) \)[/tex]: All real numbers
- Range of [tex]\( f^{-1}(x) \)[/tex]: All real numbers
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for solutions ends here at IDNLearn.com. Thank you for visiting, and come back soon for more helpful information.