Problem 94
Question
\(y=2^{x(x-1)}\) or \(x^{2}-x-\log _{2} y=0\) $$ \text { or } \quad x=\frac{1}{2}\left(1 \pm \sqrt{1+4 \log _{2} y}\right) $$ Since \(x \in[1, \infty)\), we choose \(x=\frac{1}{2}\left(1+\sqrt{1+4 \log _{2} y}\right)\) or \(\quad f^{-1}(x)=\frac{1}{2}\left(1+\sqrt{1+4 \log _{2} x}\right) .\)
Step-by-Step Solution
Verified Answer
The inverse function is \( f^{-1}(x) = \frac{1}{2}(1 + \sqrt{1 + 4 \log_2(x)}) \).
1Step 1: Recognize the Function
Identify the given functions and equations. We have the function expressed as \( y = 2^{x(x-1)} \) and another condition \( x^{2}-x- ext{log}_2(y) = 0 \). Solving these will allow us to express \( y \) in terms of \( x \).
2Step 2: Rewrite in Quadratic Form
Observe the equation \( x^2 - x - ext{log}_2(y) = 0 \), which is a quadratic equation in terms of \( x \). Recognize that we need the quadratic formula to solve for \( x \).
3Step 3: Apply the Quadratic Formula
The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -1, \text{ and } c = -\log_2(y) \) in our equation. Substituting these values, we get: \[ x = \frac{1 \pm \sqrt{1 + 4 \text{log}_2(y)}}{2} \].
4Step 4: Choose the Appropriate Solution
Because \( x \) is within the range \([1, \infty)\), choose the positive part of the quadratic solution, \( x = \frac{1}{2}\left(1 + \sqrt{1 + 4 \log_2(y)}\right) \).
5Step 5: Express the Inverse Function
Rearrange to express the inverse function \( f^{-1}(x) \), leading to the final inverse function form: \( f^{-1}(x) = \frac{1}{2}\left(1 + \sqrt{1 + 4 \log_2(x)}\right) \).
Key Concepts
Quadratic EquationsLogarithmic FunctionsFunction Transformation
Quadratic Equations
Quadratic equations play a critical role in mathematics and many fields of science. A quadratic equation is any equation that can be rearranged in standard form as \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). This type of equation represents a parabola on a graph.
In our exercise, the quadratic nature comes from the expression \( x^2 - x - \log_2(y) = 0 \). Recognizing this form helps us see that we can use the quadratic formula to solve for \( x \). The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), provides the solutions to the quadratic equation by substituting the coefficients \( a = 1 \), \( b = -1 \), and \( c = -\log_2(y) \).
Understanding and applying the quadratic formula is crucial because it provides insight into where the graph of the equation intersects the x-axis. In the context of inverse functions, solving the quadratic equation correctly is fundamental in deriving the inverse accurately.
In our exercise, the quadratic nature comes from the expression \( x^2 - x - \log_2(y) = 0 \). Recognizing this form helps us see that we can use the quadratic formula to solve for \( x \). The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), provides the solutions to the quadratic equation by substituting the coefficients \( a = 1 \), \( b = -1 \), and \( c = -\log_2(y) \).
Understanding and applying the quadratic formula is crucial because it provides insight into where the graph of the equation intersects the x-axis. In the context of inverse functions, solving the quadratic equation correctly is fundamental in deriving the inverse accurately.
Logarithmic Functions
Logarithmic functions are the inverse operations of exponential functions. They help us solve equations where the variable is in the exponent, such as \( y = 2^{x(x-1)} \). The logarithm function, specifically the base 2 logarithm \( \log_2 \), is used in the context of this exercise.
A logarithm \( \log_b(a) \) answers the question: "To what power must the base \( b \) be raised, to produce \( a \)?" In our equation, \( \log_2(y) \) signifies the power needed to raise 2 in order to achieve \( y \).
Incorporating a logarithm in a quadratic equation: \( x^2 - x - \log_2(y) = 0 \), requires an understanding of both exponential and logarithmic properties. Working with logarithms allows simplification and conversion of complex exponential expressions into more tractable forms, making it possible to apply the quadratic formula in this exercise.
A logarithm \( \log_b(a) \) answers the question: "To what power must the base \( b \) be raised, to produce \( a \)?" In our equation, \( \log_2(y) \) signifies the power needed to raise 2 in order to achieve \( y \).
Incorporating a logarithm in a quadratic equation: \( x^2 - x - \log_2(y) = 0 \), requires an understanding of both exponential and logarithmic properties. Working with logarithms allows simplification and conversion of complex exponential expressions into more tractable forms, making it possible to apply the quadratic formula in this exercise.
Function Transformation
Function transformation involves modifying the base function to achieve different shifts, stretches, or reflections. These transformations are crucial for understanding how equations and their inverses respond to various transformations.
In our solution for the inverse function, we expressed \( f^{-1}(x) = \frac{1}{2}(1 + \sqrt{1 + 4\log_2(x)}) \). This involves a shift and a root transformation of the basic quadratic setup. The original function's transformation through inverse operation provides insights into how dependent variables (like \( y \)) relate back to independent variables (like \( x \)).
Common transformations include:
In our solution for the inverse function, we expressed \( f^{-1}(x) = \frac{1}{2}(1 + \sqrt{1 + 4\log_2(x)}) \). This involves a shift and a root transformation of the basic quadratic setup. The original function's transformation through inverse operation provides insights into how dependent variables (like \( y \)) relate back to independent variables (like \( x \)).
Common transformations include:
- Horizontal and vertical shifts: moving a graph left, right, up, or down.
- Reflection: flipping a graph over a specific axis.
- Stretching and compressing: changing the graph's width or height.
Other exercises in this chapter
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