Problem 52 - Algebra and Trigonometry

Question

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 \(f^{-1} .\) \(f(x)=16-x^{2}, \quad x \geq 0\)

Step-by-Step Solution

Answer

\( f^{-1}(y) = \sqrt{16 - y} \) for \( 0 \leq y \leq 16 \).

Step 1: Understand the Function

The function given is \( f(x) = 16 - x^2 \) with \( x \geq 0 \). This is the equation of a downward opening parabola, starting from \( x = 0 \). The vertex of this parabola is at the point (0,16) since it is translated 16 units up from \( f(x) = -x^2 \) and does not move horizontally. For \( x \geq 0 \), the function is decreasing.

Step 2: Sketch the Graph of \( f \)

To graph \( f(x) = 16 - x^2 \) for \( x \geq 0 \), plot the vertex at (0,16). The parabola opens downwards because of the \( -x^2 \) term. Calculate some additional points, for example, \( f(2) = 16 - 2^2 = 12 \), gives point (2,12), \( f(4) = 16 - 4^2 = 0 \), gives point (4,0). Connect these points with a smooth curve representing the right half of the parabola.

Step 3: Sketch the Graph of \( f^{-1} \)

To sketch the inverse, reflect the graph of \( f(x) = 16 - x^2 \) over the line \( y = x \). Starting at the point (16,0), use points such as (12,2) and (0,4) from the reflection of the original graph's points (2,12) and (4,0). Connect these reflected points smoothly.

Step 4: Calculate \( f^{-1} \)

Solve \( y = 16 - x^2 \) for \( x \). Rearrange to \( x^2 = 16 - y \), and therefore \( x = \sqrt{16 - y} \) because \( x \geq 0 \). So, the inverse function is \( f^{-1}(y) = \sqrt{16 - y} \) for \( 0 \leq y \leq 16 \), matching the inversely sketched graph.

Key Concepts

Sketching Graphs

Graph sketching involves visualizing how a function behaves. Understanding features like the vertex and direction help in sketching.For the given function, understanding the form of the equation, such as \( f(x) = 16 - x^2 \), tells us the parabola opens downwards.This is due to the \(-x^2\) term, emphasizing the curve's direction.Begin by plotting key points, including the vertex, which occurs at(0,16) for this function.

• Calculate additional points by substituting \( x \) values,such as \( f(2) = 12 \), providing the point (2,12).
• Ensure accuracy by checking symmetry, given the reflectionnature of parabolas.

Connect the points with a smooth curve, clearly depicting thatthis graph only covers the range where \( x \geq 0 \). This reflectsthe function's behavior accurately for the domain given.

Parabola

A parabola is a symmetrical curve formed fromquadratic equations. The general form \( y = ax^2 + bx + c \) gives insight intohow it behaves.For our specific function \( f(x) = 16 - x^2 \), we have:

• Vertex: This is the point wherethe curve changes direction. Here, it’s at (0,16).The highest point for this downward parabola.
• Opening Direction: The curve opens downwardsas indicated by the negative \(a\) value, \(-x^2\).
• Symmetry: Parabolas are symmetricalaround their vertex. This means halves on either side ofthe vertex should mirror each other.For this function, as \( x \) increases past 0,the \( y \) values decrease symmetrically.

Knowing these features helps in correctly depicting thefunction's graph and predicting inverse behavior.

Domain and Range

In mathematics, the domain and rangedetermine all allowable input and output values.For the provided function \( f(x) = 16 - x^2 \), with\( x \geq 0 \), the domain is restricted to non-negative \( x \).

• Domain: This function's domain is\([0, \infty)\), meaning \( x \) starts from 0 and continuesinfinitely.This conforms to the portion of the parabola that is included,since only \( x \geq 0 \) adheres to the condition.
• Range: The range includes all valid\( y \) values produced by the function.For this function, the \( y \) values start at itsmaximum \( y = 16 \) at \( x = 0 \) and go down to\( y = 0 \), as \( x \) reaches 4, where the parabola intersects the \( x \)-axis.
• Inverse Function: For the inverse,\( f^{-1}(y) = \sqrt{16 - y} \), the domain becomes\([0,16]\) reflecting that y-values operate within these limits,rendering the graph's horizontal representation.

Understanding domain and range ensures correctly mapped outputsand inputs, crucial in sketching inverse graphs and recognizingfunction limits.