Answer :
Given the function [tex]\( f(x) = x^3 - 5 \)[/tex],
### Step-by-Step Solution:
1. Determine whether the function is one-to-one:
- A function is one-to-one if no horizontal line intersects its graph more than once. This can be determined by checking if its derivative is never zero over its entire domain, implying that the function is monotonic (either strictly increasing or strictly decreasing).
- Calculate the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(x^3 - 5) = 3x^2 \][/tex]
- Set the derivative equal to zero and solve:
[tex]\[ 3x^2 = 0 \implies x = 0 \][/tex]
- The derivative [tex]\( f'(x) = 3x^2 \)[/tex] is zero only at [tex]\( x = 0 \)[/tex]. Since [tex]\( 3x^2 \)[/tex] is non-negative and equals zero just at [tex]\( x = 0 \)[/tex], the function is strictly increasing everywhere except at [tex]\( x = 0 \)[/tex]. Therefore, [tex]\( f(x) = x^3 - 5 \)[/tex] is indeed a one-to-one function.
2. Finding the inverse function [tex]\( f^{-1}(x) \)[/tex]:
- To find the inverse, replace [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ y = x^3 - 5 \][/tex]
- Solve for [tex]\( x \)[/tex]:
[tex]\[ y + 5 = x^3 \implies x = (y + 5)^{1/3} \][/tex]
- Hence, the inverse function is:
[tex]\[ f^{-1}(x) = (x + 5)^{1/3} \][/tex]
Therefore, the answer is:
A. The function [tex]\( f(x) \)[/tex] is one-to-one and [tex]\( f^{-1}(x) = (x + 5)^{1/3} \)[/tex].
3. Graphing [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] on the same axes:
- The graph of [tex]\( f(x) = x^3 - 5 \)[/tex] is a cubic curve shifted downward by 5 units.
- The graph of [tex]\( f^{-1}(x) = (x + 5)^{1/3} \)[/tex] can be obtained by reflecting the graph of [tex]\( f(x) \)[/tex] across the line [tex]\( y = x \)[/tex].
Typically, you would sketch [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex] on the same set of axes, showing their symmetry about the line [tex]\( y = x \)[/tex].
4. Domain and Range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]:
- The domain of [tex]\( f(x) = x^3 - 5 \)[/tex] is all real numbers:
[tex]\[ \text{Domain of } f: (-\infty, \infty) \][/tex]
- The range of [tex]\( f(x) = x^3 - 5 \)[/tex] is also all real numbers:
[tex]\[ \text{Range of } f: (-\infty, \infty) \][/tex]
- The domain of [tex]\( f^{-1}(x) = (x + 5)^{1/3} \)[/tex] is all real numbers:
[tex]\[ \text{Domain of } f^{-1}: (-\infty, \infty) \][/tex]
- The range of [tex]\( f^{-1}(x) = (x + 5)^{1/3} \)[/tex] is also all real numbers:
[tex]\[ \text{Range of } f^{-1}: (-\infty, \infty) \][/tex]
In conclusion:
- The function [tex]\( f(x) = x^3 - 5 \)[/tex] is one-to-one.
- The inverse function is [tex]\( f^{-1}(x) = (x + 5)^{1/3} \)[/tex].
- The domain and range for both [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] are all real numbers.
### Step-by-Step Solution:
1. Determine whether the function is one-to-one:
- A function is one-to-one if no horizontal line intersects its graph more than once. This can be determined by checking if its derivative is never zero over its entire domain, implying that the function is monotonic (either strictly increasing or strictly decreasing).
- Calculate the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(x^3 - 5) = 3x^2 \][/tex]
- Set the derivative equal to zero and solve:
[tex]\[ 3x^2 = 0 \implies x = 0 \][/tex]
- The derivative [tex]\( f'(x) = 3x^2 \)[/tex] is zero only at [tex]\( x = 0 \)[/tex]. Since [tex]\( 3x^2 \)[/tex] is non-negative and equals zero just at [tex]\( x = 0 \)[/tex], the function is strictly increasing everywhere except at [tex]\( x = 0 \)[/tex]. Therefore, [tex]\( f(x) = x^3 - 5 \)[/tex] is indeed a one-to-one function.
2. Finding the inverse function [tex]\( f^{-1}(x) \)[/tex]:
- To find the inverse, replace [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ y = x^3 - 5 \][/tex]
- Solve for [tex]\( x \)[/tex]:
[tex]\[ y + 5 = x^3 \implies x = (y + 5)^{1/3} \][/tex]
- Hence, the inverse function is:
[tex]\[ f^{-1}(x) = (x + 5)^{1/3} \][/tex]
Therefore, the answer is:
A. The function [tex]\( f(x) \)[/tex] is one-to-one and [tex]\( f^{-1}(x) = (x + 5)^{1/3} \)[/tex].
3. Graphing [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] on the same axes:
- The graph of [tex]\( f(x) = x^3 - 5 \)[/tex] is a cubic curve shifted downward by 5 units.
- The graph of [tex]\( f^{-1}(x) = (x + 5)^{1/3} \)[/tex] can be obtained by reflecting the graph of [tex]\( f(x) \)[/tex] across the line [tex]\( y = x \)[/tex].
Typically, you would sketch [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex] on the same set of axes, showing their symmetry about the line [tex]\( y = x \)[/tex].
4. Domain and Range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]:
- The domain of [tex]\( f(x) = x^3 - 5 \)[/tex] is all real numbers:
[tex]\[ \text{Domain of } f: (-\infty, \infty) \][/tex]
- The range of [tex]\( f(x) = x^3 - 5 \)[/tex] is also all real numbers:
[tex]\[ \text{Range of } f: (-\infty, \infty) \][/tex]
- The domain of [tex]\( f^{-1}(x) = (x + 5)^{1/3} \)[/tex] is all real numbers:
[tex]\[ \text{Domain of } f^{-1}: (-\infty, \infty) \][/tex]
- The range of [tex]\( f^{-1}(x) = (x + 5)^{1/3} \)[/tex] is also all real numbers:
[tex]\[ \text{Range of } f^{-1}: (-\infty, \infty) \][/tex]
In conclusion:
- The function [tex]\( f(x) = x^3 - 5 \)[/tex] is one-to-one.
- The inverse function is [tex]\( f^{-1}(x) = (x + 5)^{1/3} \)[/tex].
- The domain and range for both [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] are all real numbers.
