The Function F ( X ) = 5 X + 4 F(x) = 5x + 4 F ( X ) = 5 X + 4 Is One-to-one.a. Find An Equation For F − 1 F^{-1} F − 1 , The Inverse Function.b. Verify That Your Equation Is Correct By Showing That F ( F − 1 ( X ) ) = X F(f^{-1}(x)) = X F ( F − 1 ( X )) = X And F − 1 ( F ( X ) ) = X F^{-1}(f(x)) = X F − 1 ( F ( X )) = X .

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Introduction

In mathematics, a one-to-one function is a function that maps each element of its domain to a unique element in its range. In other words, no two elements in the domain of a one-to-one function can map to the same element in the range. The function $f(x) = 5x + 4$ is a linear function that is one-to-one, meaning that each value of $x$ maps to a unique value of $f(x)$. In this article, we will find an equation for $f^{-1}$, the inverse function of $f(x) = 5x + 4$, and verify that our equation is correct by showing that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.

Finding the Inverse Function

To find the inverse function of $f(x) = 5x + 4$, we need to swap the roles of $x$ and $f(x)$ and solve for $f(x)$. Let $y = 5x + 4$. To find the inverse function, we need to solve for $x$ in terms of $y$.

Step 1: Swap the Roles of $x$ and $f(x)$

We start by swapping the roles of $x$ and $f(x)$, which gives us $x = 5y + 4$.

Step 2: Solve for $y$

Now, we need to solve for $y$ in terms of $x$. To do this, we can isolate $y$ on one side of the equation.

# Import necessary modules
import sympy as sp
x, y = sp.symbols('x y')

eq = sp.Eq(x, 5*y + 4)

y_sol = sp.solve(eq, y)[0]
print(y_sol)

The output of the code above is:

$$y = \frac{x - 4}{5}$$

Step 3: Write the Inverse Function

Now that we have solved for $y$ in terms of $x$, we can write the inverse function of $f(x) = 5x + 4$ as $f^{-1}(x) = \frac{x - 4}{5}$.

Verifying the Inverse Function

To verify that our equation for $f^{-1}$ is correct, we need to show that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.

Step 1: Show that $f(f^{-1}(x)) = x$

We start by substituting $f^{-1}(x) = \frac{x - 4}{5}$ into the equation for $f(x) = 5x + 4$.

# Import necessary modules
import sympy as sp

x = sp.symbols('x')

f_x = 5*x + 4

f_inv_x = (x - 4)/5

f_f_inv_x = f_x.subs(x, f_inv_x)
print(f_f_inv_x)

The output of the code above is:

$$x$$

This shows that $f(f^{-1}(x)) = x$, which verifies that our equation for $f^{-1}$ is correct.

Step 2: Show that $f^{-1}(f(x)) = x$

We start by substituting $f(x) = 5x + 4$ into the equation for $f^{-1}(x) = \frac{x - 4}{5}$.

# Import necessary modules
import sympy as sp

x = sp.symbols('x')

f_x = 5*x + 4

f_inv_x = (x - 4)/5

f_inv_f_x = f_inv_x.subs(x, f_x)
print(f_inv_f_x)

The output of the code above is:

$$x$$

This shows that $f^{-1}(f(x)) = x$, which verifies that our equation for $f^{-1}$ is correct.

Conclusion

In this article, we found an equation for $f^{-1}$, the inverse function of $f(x) = 5x + 4$, and verified that our equation is correct by showing that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. The inverse function of $f(x) = 5x + 4$ is given by $f^{-1}(x) = \frac{x - 4}{5}$.

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Introduction

In our previous article, we found an equation for $f^{-1}$, the inverse function of $f(x) = 5x + 4$, and verified that our equation is correct by showing that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. In this article, we will answer some common questions related to the function $f(x) = 5x + 4$ and its inverse function $f^{-1}(x) = \frac{x - 4}{5}$.

Q&A

Q: What is the domain and range of the function $f(x) = 5x + 4$?

A: The domain of the function $f(x) = 5x + 4$ is all real numbers, and the range is also all real numbers.

Q: What is the domain and range of the inverse function $f^{-1}(x) = \frac{x - 4}{5}$?

A: The domain of the inverse function $f^{-1}(x) = \frac{x - 4}{5}$ is all real numbers, and the range is also all real numbers.

Q: How do you know that the function $f(x) = 5x + 4$ is one-to-one?

A: The function $f(x) = 5x + 4$ is one-to-one because it passes the horizontal line test. This means that no horizontal line intersects the graph of the function more than once.

Q: How do you find the inverse function of a one-to-one function?

A: To find the inverse function of a one-to-one function, you need to swap the roles of $x$ and $f(x)$ and solve for $f(x)$. This is done by replacing $f(x)$ with $y$ and then solving for $y$ in terms of $x$.

Q: What is the relationship between the function $f(x) = 5x + 4$ and its inverse function $f^{-1}(x) = \frac{x - 4}{5}$?

A: The function $f(x) = 5x + 4$ and its inverse function $f^{-1}(x) = \frac{x - 4}{5}$ are inverse functions of each other. This means that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.

Q: How do you verify that the inverse function of a one-to-one function is correct?

A: To verify that the inverse function of a one-to-one function is correct, you need to show that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. This can be done by substituting the inverse function into the original function and simplifying.

Q: What is the significance of the inverse function in real-world applications?

A: The inverse function is significant in real-world applications because it allows us to solve equations and find the values of unknown variables. For example, if we have an equation $y = 5x + 4$, we can use the inverse function to find the value of $x$ in terms of $y$.

Conclusion

In this article, we answered some common questions related to the function $f(x) = 5x + 4$ and its inverse function $f^{-1}(x) = \frac{x - 4}{5}$. We hope that this article has provided you with a better understanding of the concept of inverse functions and how they are used in real-world applications.

Additional Resources

• Inverse Functions
• One-to-One Functions
• Horizontal Line Test

Final Thoughts

The concept of inverse functions is an important one in mathematics, and it has many real-world applications. By understanding how to find the inverse function of a one-to-one function, we can solve equations and find the values of unknown variables. We hope that this article has provided you with a better understanding of the concept of inverse functions and how they are used in real-world applications.