Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
To determine whether the function [tex]\( f(x) = 4x - 7 \)[/tex] is one-to-one, we need to examine its properties:
1. Is [tex]\( f(x) \)[/tex] one-to-one?
A function is one-to-one if it never takes the same value twice; that is, [tex]\( f(a) = f(b) \)[/tex] implies [tex]\( a = b \)[/tex]. Another way to determine if a function is one-to-one is to check its derivative. If the derivative is always positive or always negative, the function is strictly monotonic and hence one-to-one.
The derivative of [tex]\( f(x) = 4x - 7 \)[/tex] is [tex]\( f'(x) = 4 \)[/tex]. Since this derivative is a positive constant, [tex]\( f(x) \)[/tex] is always increasing. Therefore, [tex]\( f(x) \)[/tex] is one-to-one.
2. (a) Write an equation for the inverse function in the form [tex]\( y = f^{-1}(x) \)[/tex]:
To find the inverse function, we need to solve the equation [tex]\( y = 4x - 7 \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ \begin{align*} y + 7 &= 4x \\ x &= \frac{y + 7}{4} \end{align*} \][/tex]
Thus, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x + 7}{4} \][/tex]
3. (b) Graphing [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]:
While graphing here isn't practical, you can visualize that both [tex]\( f(x) = 4x - 7 \)[/tex] and [tex]\( f^{-1}(x) = \frac{x + 7}{4} \)[/tex] are lines. Specifically:
- [tex]\( f(x) \)[/tex] is a line with a slope of 4 and a y-intercept at -7.
- [tex]\( f^{-1}(x) \)[/tex] is a line with a slope of [tex]\( \frac{1}{4} \)[/tex] and a y-intercept at [tex]\( \frac{7}{4} \)[/tex].
On the coordinate plane, these lines are mirror images of each other over the line [tex]\( y = x \)[/tex].
4. (c) Domain and Range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]:
- For the function [tex]\( f(x) = 4x - 7 \)[/tex]:
- Domain: All real numbers ([tex]\(\mathbb{R}\)[/tex])
- Range: All real numbers ([tex]\(\mathbb{R}\)[/tex])
- For the inverse function [tex]\( f^{-1}(x) = \frac{x + 7}{4} \)[/tex]:
- Domain: All real numbers ([tex]\(\mathbb{R}\)[/tex])
- Range: All real numbers ([tex]\(\mathbb{R}\)[/tex])
Since both functions are linear and their slopes are non-zero, their domains and ranges cover all real numbers.
Summary:
- The function [tex]\( f(x) = 4x - 7 \)[/tex] is one-to-one.
- The inverse function is [tex]\( f^{-1}(x) = \frac{x + 7}{4} \)[/tex].
- The domain and range of both [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex] are all real numbers ([tex]\(\mathbb{R}\)[/tex]).
1. Is [tex]\( f(x) \)[/tex] one-to-one?
A function is one-to-one if it never takes the same value twice; that is, [tex]\( f(a) = f(b) \)[/tex] implies [tex]\( a = b \)[/tex]. Another way to determine if a function is one-to-one is to check its derivative. If the derivative is always positive or always negative, the function is strictly monotonic and hence one-to-one.
The derivative of [tex]\( f(x) = 4x - 7 \)[/tex] is [tex]\( f'(x) = 4 \)[/tex]. Since this derivative is a positive constant, [tex]\( f(x) \)[/tex] is always increasing. Therefore, [tex]\( f(x) \)[/tex] is one-to-one.
2. (a) Write an equation for the inverse function in the form [tex]\( y = f^{-1}(x) \)[/tex]:
To find the inverse function, we need to solve the equation [tex]\( y = 4x - 7 \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ \begin{align*} y + 7 &= 4x \\ x &= \frac{y + 7}{4} \end{align*} \][/tex]
Thus, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x + 7}{4} \][/tex]
3. (b) Graphing [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]:
While graphing here isn't practical, you can visualize that both [tex]\( f(x) = 4x - 7 \)[/tex] and [tex]\( f^{-1}(x) = \frac{x + 7}{4} \)[/tex] are lines. Specifically:
- [tex]\( f(x) \)[/tex] is a line with a slope of 4 and a y-intercept at -7.
- [tex]\( f^{-1}(x) \)[/tex] is a line with a slope of [tex]\( \frac{1}{4} \)[/tex] and a y-intercept at [tex]\( \frac{7}{4} \)[/tex].
On the coordinate plane, these lines are mirror images of each other over the line [tex]\( y = x \)[/tex].
4. (c) Domain and Range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]:
- For the function [tex]\( f(x) = 4x - 7 \)[/tex]:
- Domain: All real numbers ([tex]\(\mathbb{R}\)[/tex])
- Range: All real numbers ([tex]\(\mathbb{R}\)[/tex])
- For the inverse function [tex]\( f^{-1}(x) = \frac{x + 7}{4} \)[/tex]:
- Domain: All real numbers ([tex]\(\mathbb{R}\)[/tex])
- Range: All real numbers ([tex]\(\mathbb{R}\)[/tex])
Since both functions are linear and their slopes are non-zero, their domains and ranges cover all real numbers.
Summary:
- The function [tex]\( f(x) = 4x - 7 \)[/tex] is one-to-one.
- The inverse function is [tex]\( f^{-1}(x) = \frac{x + 7}{4} \)[/tex].
- The domain and range of both [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex] are all real numbers ([tex]\(\mathbb{R}\)[/tex]).
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.