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Magazine Chapter 5: Problem 12
Show that the given function is one-to-one and find its inverse. Check your answers algebraically and graphically. Verify that the range of \(f\) is the domain of \(f^{-1}\) and vice-versa. $$f(x)=3(x+4)^{2}-5, x \leq-4$$
Short Answer
Expert verified
The function is one-to-one for \( x \leq -4 \), the inverse is \( f^{-1}(x) = -4 - \sqrt{\frac{x+5}{3}} \), and both algebraic and graphical checks confirm correctness.
Step by step solution
01
Introduction to the Function
We begin by understanding the function defined as \( f(x) = 3(x+4)^2 - 5 \) with the domain restriction \( x \leq -4 \). The goal is to show that this function is one-to-one and then find its inverse.
02
Determine One-to-One Function
Since the function is a quadratic in the form of \( f(x) = a(x - h)^2 + k \) and \( a > 0 \), the parabola opens upwards. To ensure it is one-to-one, we analyze it within the specified domain \( x \leq -4 \). Observing this, the function is monotonically increasing when considering the vertex \( x = -4 \) as the lowest point within our specified domain. Hence, \( f(x) \) is one-to-one for \( x \leq -4 \).
03
Express \( y \) in terms of \( x \) for the Inverse
Set \( y = f(x) = 3(x+4)^2 - 5 \). Now, solve for \( x \) in terms of \( y \).
04
Solve for \( x \)
First, isolate the quadratic term: \( y + 5 = 3(x+4)^2 \). Divide by 3 to get \( \frac{y+5}{3} = (x+4)^2 \). Take the square root on both sides, remembering to consider the negative root due to domain restriction: \( x+4 = -\sqrt{\frac{y+5}{3}} \). Then solve for \( x \): \( x = -4 - \sqrt{\frac{y+5}{3}} \).
05
Find the Inverse Function
Replace \( y \) with \( x \) to express the inverse function: \( f^{-1}(x) = -4 - \sqrt{\frac{x+5}{3}} \). This is the inverse of \( f \) within the given domain constraints.
06
Verify Algebraically
Check that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \). For \( f(f^{-1}(x)) \): replace \( y \) with \( f^{-1}(x) \) in \( f \), showing it simplifies to \( x \). For \( f^{-1}(f(x)) \): replace \( f(x) \) with \( y \) in \( f^{-1} \), simplifying back to \( x \). Both operations prove the inverse relationship.
07
Graph Verification
Graph both \( f(x) \) and \( f^{-1}(x) \) on the same set of axes. We expect the graphs to be reflections over the line \( y = x \). The graphical check verifies correctness if the shapes and intersection meet these criteria.
08
Verify Domains and Ranges
The original function has a domain of \( x \leq -4 \), corresponding to a range of \( y \geq -5 \). The inverse function should therefore have a domain of \( x \geq -5 \) and a range of \( y \leq -4 \), which matches our results.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
One-to-One Functions
A function is termed **one-to-one** if it never assigns the same value to different elements of its domain. This implies that each input is mapped to a unique output, making it possible to define an inverse function.
For a given quadratic function, even though it usually has a symmetrical curve and fails the one-to-one test, we can still demonstrate that it is one-to-one over a restricted domain.
For a given quadratic function, even though it usually has a symmetrical curve and fails the one-to-one test, we can still demonstrate that it is one-to-one over a restricted domain.
- Consider the function \( f(x) = 3(x+4)^2 - 5 \) with a domain \( x \leq -4 \).
- By focusing on this domain, the function behaves as a monotonic upward trend starting from the vertex at \( x = -4 \).
- This monotonic increase ensures each \( x \) maps to a distinct \( y \), making \( f(x) \) one-to-one within the restricted domain.
Quadratic Functions
Quadratic functions are polynomial functions with the degree of two, generally taking the form \( f(x) = ax^2 + bx + c \). They are typically represented graphically by parabolas. Here's more about how they work:
- A parabola's direction of opening, upward or downward, is determined by the sign of \( a \). If \( a > 0 \), it opens upwards; if \( a < 0 \), it opens downwards.
- In our specific function, \( f(x) = 3(x+4)^2 - 5 \), since \( a = 3 > 0 \), the parabola opens upwards.
- The vertex represents the turning point of the parabola. For our exercise, it occurs at \( x = -4 \), showing that above this point, the function increases, supporting its behavior as one-to-one within the given domain \( x \leq -4 \).
Domain and Range
The domain and range are fundamental concepts that specify limits on what values a function can take. For the function \( f(x) = 3(x+4)^2 - 5 \), these concepts are critical.
- The domain specifies the inputs available for a function. Here, the domain is \( x \leq -4 \) due to a restriction ensuring the one-to-one nature of the function.
- Based on this domain, the smallest possible value of \( f(x) \) occurs at \( x = -4 \), which is \( f(-4) = -5 \). Therefore, the range is \( y \geq -5 \).
- The inverse function inverts the domain and range, meaning the range of the original function (\( y \geq -5 \)) becomes the domain of the inverse (\( x \geq -5 \)), and vice-versa for the inverse's range (\( y \leq -4 \)).
Graphical Verification
Graphical verification is a method to visually confirm the properties of functions and their inverses.
- For any function \( f(x) \) and its inverse \( f^{-1}(x) \), plotting both graphs can help verify the inverse relationship.
- The graphs should reflect over the line \( y = x \), meaning if you take a point from the graph of \( f(x) \) and switch its coordinates, you should find it on the graph of \( f^{-1}(x) \).
- In our exercise, drawing \( f(x) = 3(x+4)^2 - 5 \) and its inverse \( f^{-1}(x) = -4 - \sqrt{\frac{x+5}{3}} \) should yield reflected paths over \( y = x \).
