Question

Bilden Sie die Umkehrfunktion \(\mathrm{y}^{*}\) zu folgenden Funktionen a) \(y=2 x-5\) b) \(y=8 x^{3}+1\) c) \(y=\ln 2 x\)

Short answer

a) y* = (x + 5)/2; b) y* = ((x-1)/8)^(1/3); c) y* = e^x / 2

Step-by-step solution

Identify the Original Function - Part a

Given the function: y = 2x - 5

Rewrite the Function in Terms of x

Swap y and x and solve for the new y to find the inverse: x = 2y - 5

Solve for y

Solve for y in terms of x: x + 5 = 2y y = (x + 5)/2

Final Inverse Function - Part a

The inverse function for part a is y* = (x + 5)/2

Identify the Original Function - Part b

Given the function: y = 8x^3 + 1

Rewrite the Function in Terms of x

Swap y and x and solve for the new y to find the inverse: x = 8y^3 + 1

Solve for y

Solve for y in terms of x: x - 1 = 8y^3 y^3 = (x - 1)/8 y = ((x - 1)/8)^(1/3)

Final Inverse Function - Part b

The inverse function for part b is y* = ((x - 1)/8)^(1/3)

Identify the Original Function - Part c

Given the function: y = ln(2x)

Rewrite the Function in Terms of x

Swap y and x and solve for the new y to find the inverse: x = ln(2y)

Solve for y

Solve for y in terms of x: e^x = 2y y = e^x / 2

Final Inverse Function - Part c

The inverse function for part c is y* = e^x / 2

Key concepts

Algebra

In algebra, we deal with variables and constants to formulate equations. The process of solving algebraic equations involves finding the values of unknown variables. When finding the inverse of a function, we essentially reverse the operations performed on the variable to get the original function. For instance, if an equation involves adding 5 to a variable and then multiplying by 2, the inverse operation would involve dividing by 2 and then subtracting 5.
Let's look at an example:

• Original function: \(y = 2x - 5\)
• Inverse function: \(y^* = \frac{x + 5}{2}\)

Understanding these operations and their reversals is fundamental in algebra and essential when dealing with inverse functions.

Functions

A function is a mathematical relationship between two variables, typically denoted as x and y, where each input (x) corresponds to one output (y). Inverse functions are a special type of function that essentially 'undo' the operation of the original function. If the function is \(y = f(x)\), its inverse is denoted as \(y^* = f^{-1}(x)\). Finding the inverse involves swapping the roles of x and y and solving for the new y.
Here's a breakdown of the process:

• Original function: \(y = 8x^3 + 1\)
• Swap x and y: \(x = 8y^3 + 1\)
• Solve for y: \(y = \sqrt[3]{\frac{x - 1}{8}}\)

This concept is crucial for understanding more complex functional relationships and how to reverse them in various mathematical problems.

Logarithms

Logarithms are the inverse of exponential functions. They answer the question: 'To what power must a base be raised to obtain a certain value?' For example, the logarithm base 2 of 8 is 3, because \(2^3 = 8\). The natural logarithm (ln) has a base e (approximately 2.718). When dealing with inverse functions involving logarithms, we follow similar steps as before.
Consider the function \(y = \ln(2x)\):

• Original function: \(y = \ln(2x)\)

Swap x and y: \(x = \ln(2y)\)
Solve for y: \(2y = e^x\)

Final inverse: \(y = \frac{e^x}{2}\)

Logarithms simplify many complex multiplicative relationships into more manageable additive ones, making them powerful tools in mathematics.

Cubic Functions

Cubic functions are polynomial equations of the form \(y = ax^3 + bx^2 + cx + d\). They produce more complex curved graphs than linear or quadratic functions. When finding the inverse of a cubic function, the process might seem daunting due to the nature of cubed terms. However, it involves similar steps of reversing operations.
Consider the function \(y = 8x^3 + 1\):

• Original function: \(y = 8x^3 + 1\)
• Swap x and y: \(x = 8y^3 + 1\)
• Solve for y: \(y = \sqrt[3]{\frac{x - 1}{8}}\)

Understanding how to handle these higher-order terms is key in algebra and calculus, where such functions frequently appear. By mastering this, students can tackle a wide range of problems involving polynomial equations.