Chapter 2: Problem 52
A function \(f\) is given. (a) Sketch the graph of \(f\) (b) Use the graph of \(f\) to sketch the graph of \(f^{-1}\) (c) Find \(f^{-1}\) $$f(x)=16-x^{2}, \quad x \geq 0$$
Short Answer
Expert verified
(c) The inverse is \(f^{-1}(x)=\sqrt{16-x}, \; x \leq 16\).
Step by step solution
01
Understand the function
The given function is \( f(x) = 16 - x^2 \) with the restriction \( x \geq 0 \). This means it's a downward-opening parabola starting from the x-axis (at \( x=0 \)) and reflects only the right half due to the domain restriction.
02
Sketch the graph of $f(x)$
The vertex of the parabola is at the point \((0, 16)\), and it opens downward. Since \( x \geq 0 \), we only draw the right side of the parabola from the vertex.
03
Reflect on the y=x line to find $f^{-1}(x)$
To sketch the inverse \( f^{-1} \), we reflect the graph of \( f(x) \) with respect to the line \( y = x \). For the function \( y = 16 - x^2 \), swap \( x \) and \( y \), resulting in the graph of \( x = 16 - y^2 \) for the inverse.
04
Solve for $f^{-1}(x)$ algebraically
Start by swapping \( x \) and \( y \) in the equation to obtain \( x = 16 - y^2 \). Solve for \( y \) when \( x \geq 0 \):1. \( y^2 = 16 - x \).2. \( y = \sqrt{16 - x} \). Thus, \( f^{-1}(x) = \sqrt{16 - x} \), where \( x \leq 16 \).
05
Verify the domain and range
The function \( f(x) \) has its domain restricted to \( x \geq 0 \) and the range \( 0 \leq y \leq 16 \). The inverse function \( f^{-1}(x) = \sqrt{16 - x} \) thus has domain \( 0 \leq x \leq 16 \) and range \( y \geq 0 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Graph Sketching
When sketching graphs, it's like painting a picture of an equation. In our case, we have a downward-opening parabola described by the function \( f(x) = 16 - x^2 \). This indicates that the curve opens downward, and the highest point, or vertex, is at \((0, 16)\).
Imagine drawing a smiley face, but with only half a curve because of the condition \( x \geq 0 \). This means you only draw the right half, starting from the vertex straight down. The more detail, the more informative your graph.
Sketching the inverse function \( f^{-1} \) involves a reflection over the line \( y = x \). This can be thought of as flipping the graph so that every point \((a, b)\) becomes \((b, a)\). It’s like looking into a mirror and swapping your x and y coordinates.
So, by accurately sketching both the function and its inverse, you gain a complete visual understanding of their relationship.
Imagine drawing a smiley face, but with only half a curve because of the condition \( x \geq 0 \). This means you only draw the right half, starting from the vertex straight down. The more detail, the more informative your graph.
Sketching the inverse function \( f^{-1} \) involves a reflection over the line \( y = x \). This can be thought of as flipping the graph so that every point \((a, b)\) becomes \((b, a)\). It’s like looking into a mirror and swapping your x and y coordinates.
So, by accurately sketching both the function and its inverse, you gain a complete visual understanding of their relationship.
Domain and Range
The domain and range of a function are essential to define its scope and reach. For our parabolic function, \( f(x) = 16 - x^2 \), we have a specific domain: \( x \geq 0 \). This tells us that we're only considering the right half of the parabola.
The range for this function is \(0 \leq y \leq 16\), as the parabola peaks at 16 and extends downward. It's the full set of potential outputs or y-values the function can achieve.
Now, consider the inverse function \( f^{-1}(x) \). The domain of the original becomes the range of the inverse and vice versa. Thus, \( f^{-1}(x) = \sqrt{16 - x} \) has a domain \(0 \leq x \leq 16\) and range \(y \geq 0\). Understanding and defining these domains and ranges is crucial when working with inverse functions.
The range for this function is \(0 \leq y \leq 16\), as the parabola peaks at 16 and extends downward. It's the full set of potential outputs or y-values the function can achieve.
Now, consider the inverse function \( f^{-1}(x) \). The domain of the original becomes the range of the inverse and vice versa. Thus, \( f^{-1}(x) = \sqrt{16 - x} \) has a domain \(0 \leq x \leq 16\) and range \(y \geq 0\). Understanding and defining these domains and ranges is crucial when working with inverse functions.
- Original Function: \( x \geq 0 \) (Domain), \(0 \leq y \leq 16\) (Range)
- Inverse Function: \(0 \leq x \leq 16\) (Domain), \(y \geq 0\) (Range)
Parabolas
Parabolas are fascinating structures in algebra and geometry. At their essence, parabolas form symmetrical, U-shaped curves.
Our equation \( f(x) = 16 - x^2 \) defines a downward-opening parabola. The negative coefficient before \(x^2\) signals this downward opening, like a frown.
Each parabola has a key feature: the vertex. This is the point where the parabola changes direction. For our equation, the vertex is at \((0, 16)\), since without any additions to \(x\), its y-intercept is also its vertex.
Graphing parabolas requires observing their properties:
Our equation \( f(x) = 16 - x^2 \) defines a downward-opening parabola. The negative coefficient before \(x^2\) signals this downward opening, like a frown.
Each parabola has a key feature: the vertex. This is the point where the parabola changes direction. For our equation, the vertex is at \((0, 16)\), since without any additions to \(x\), its y-intercept is also its vertex.
Graphing parabolas requires observing their properties:
- Axis of symmetry: This is the vertical line that divides the parabola into mirror images. Here, it’s the y-axis or \(x = 0\).
- Vertex: The highest or lowest point, depending on the parabola’s direction. In our case, it’s the highest point.
- Direction: Determined by the sign of the \(x^2\) term. Negative means down.