Chapter 4: Problem 29
Find a formula for \(f^{-1}(x),\) and state the domain of \(f^{-1}\). $$f(x)=x-5 x^{2}, x \geq 1$$
Short Answer
Expert verified
\( f^{-1}(x) = \frac{1 + \sqrt{1-20x}}{10} \); domain: \( x \in [0, \frac{1}{20}] \).
Step by step solution
01
Understand the Problem
We need to find the inverse of the function \( f(x) = x - 5x^2 \), where \( x \geq 1 \). We also need to determine the domain of \( f^{-1}(x) \).
02
Replace the Function
To find the inverse, start by letting \( y = f(x) \). Thus, \( y = x - 5x^2 \).
03
Solve for x in terms of y
We need to express \( x \) in terms of \( y \). Rearrange the equation \( y = x - 5x^2 \) to the form of a quadratic equation: \( 5x^2 - x + y = 0 \).
04
Use the Quadratic Formula
The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) will be utilized to solve for \( x \). Here, \( a = 5 \), \( b = -1 \), and \( c = y \). The solution is: \[ x = \frac{1 \pm \sqrt{1 - 20y}}{10} \].
05
Determine the Correct Root
Since \( x \geq 1 \), use the positive root to ensure nonsensical values like negative x-values are not considered. So, \( x = \frac{1 + \sqrt{1-20y}}{10} \).
06
Write the Inverse Function
The inverse function is \( f^{-1}(x) = \frac{1 + \sqrt{1-20x}}{10} \).
07
Determine the Domain of the Inverse Function
For the inverse to exist, the expression under the square root \( 1 - 20x \) should be non-negative. This means \( x \leq \frac{1}{20} \). Therefore, the domain of \( f^{-1}(x) \) is \( x \in [0, \frac{1}{20}] \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equations
A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). This type of equation is characterized by the highest exponent of the variable, usually \( x \), being 2. Quadratics are pivotal in determining the roots or solutions where the function intersects the x-axis.
To solve a quadratic, we can use various methods:
This equation format allowed us to express the inverse function’s formula after rearranging the original function. By recognizing the quadratic form of \( 5x^2 - x + y = 0 \), we realized it was solvable using the quadratic formula, which provided the bridge to isolating \( x \) for the inverse.
To solve a quadratic, we can use various methods:
- The quadratic formula, given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
- Factoring, if the quadratic can be expressed as a product of linear factors.
- Completing the square, which involves transforming a quadratic into a perfect square trinomial.
This equation format allowed us to express the inverse function’s formula after rearranging the original function. By recognizing the quadratic form of \( 5x^2 - x + y = 0 \), we realized it was solvable using the quadratic formula, which provided the bridge to isolating \( x \) for the inverse.
Domain of a Function
The domain of a function is the complete set of possible values of the independent variable (often \( x \)) for which the function is defined. It's important as it determines the range over which the function can operate without generating undefined or nonsensical outputs. For a function's inverse, the domain plays a crucial role because it signifies the values for which the inverse remains valid.
When we find the inverse of a function, its domain is derived based on the original function's range. In our case, to ensure the expression under the square root, \( 1 - 20x \), remains non-negative, we need \( x \leq \frac{1}{20} \). Thus, the domain of \( f^{-1}(x) \) is \( x \in [0, \frac{1}{20}] \), ensuring all values work in the inverse's formula.
When we find the inverse of a function, its domain is derived based on the original function's range. In our case, to ensure the expression under the square root, \( 1 - 20x \), remains non-negative, we need \( x \leq \frac{1}{20} \). Thus, the domain of \( f^{-1}(x) \) is \( x \in [0, \frac{1}{20}] \), ensuring all values work in the inverse's formula.
Square Root
The square root is a mathematical operation that finds a number which, when multiplied by itself, results in the original number. For example, the square root of 16 is 4, because \( 4 \times 4 = 16 \). A vital property of the square root operation is that it returns non-negative results when applied to non-negative inputs.
The function \( \sqrt{x} \) is defined only when \( x \geq 0 \), making it significant in determining the domain of functions involving square roots.
In dealing with inverse functions, ensuring the expression inside the square root is non-negative is key. In our exercise, \( \sqrt{1-20x} \) needed to exist only for \( 1-20x \geq 0 \). Respecting this constraint allowed us to find the correct domain of the inverse function, ensuring all mathematical operations within it were valid and real.
The function \( \sqrt{x} \) is defined only when \( x \geq 0 \), making it significant in determining the domain of functions involving square roots.
In dealing with inverse functions, ensuring the expression inside the square root is non-negative is key. In our exercise, \( \sqrt{1-20x} \) needed to exist only for \( 1-20x \geq 0 \). Respecting this constraint allowed us to find the correct domain of the inverse function, ensuring all mathematical operations within it were valid and real.