Problem 62 - 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

The inverse function is \(f^{-1}(x) = \sqrt{16 - x}\) for \(0 \leq x \leq 16\).

Step 1: Analyze the Function

The function given is \(f(x) = 16 - x^2\) with \(x \geq 0\). It's a downward opening parabola because the coefficient of \(x^2\) is negative. Since \(x \geq 0\), consider only the right half of the parabola from its vertex at \(x = 0 \) to the positive x-axis.

Step 2: Sketch the Graph of f

Plot the function \(f(x) = 16 - x^2\) over its domain (\(x \geq 0\)). Start at the vertex \((0, 16)\) and make the parabola open downwards towards \(x = 4\), since at \(x = 4, f(x) = 0\). This captures the portion where \(x \geq 0\).

Step 3: Understand the Inverse Function Conceptually

The inverse function, \(f^{-1}(x)\), will essentially switch the roles of \(x\) and \(y\) in the function \(f(x)\). To find the inverse, solve \(y = 16 - x^2\) for \(x\).

Step 4: Solve for the Inverse Function

Set \(y = 16 - x^2\) and rearrange to get \(x^2 = 16 - y\). Since \(x \geq 0\), we take the positive square root to solve for \(x\). Thus, \(x = \sqrt{16 - y}\). Consequently, the inverse function is \(f^{-1}(x) = \sqrt{16 - x}\) for \(0 \leq x \leq 16\).

Step 5: Sketch the Graph of f^{-1}

Reflect the graph of \(f\) over the line \(y = x\). This is done by plotting the points where \(f(x) = y\) as \((y, x)\). Given \(f(x) = 16 - x^2\) for \(x \geq 0\), inverse relationships like \((0, 16) \rightarrow (16, 0)\), and \((4, 0) \rightarrow (0, 4)\) hold for \(f^{-1}\). Sketch \(f^{-1}(x) = \sqrt{16 - x}\) from \(x = 0\) to \(x = 16\).

Key Concepts

Sketching Graphs

Starting with sketching the graph of a function might seem daunting, but it's all about finding and connecting key points. For a given function, like \(f(x) = 16 - x^2\) with the domain restriction \(x \geq 0\), you focus on certain markers:

• The vertex of the parabola, which here is at \((0, 16)\).
• The points where the function meets the axes, giving insights into the shape and direction of the graph.

You begin at the vertex and plot points such as \((1, 15)\) and so forth, until the function reaches \((4, 0)\).
Connect these dots smoothly to maintain the parabolic shape.
Visualizing the graph supports understanding of how the function behaves over the specified domain.

Parabolas

Parabolas are U-shaped graphs that can either open upwards or downwards. For the function \(f(x) = 16 - x^2\), we have a downward-opening parabola. The negative sign before \(x^2\) indicates this downward opening.
Another feature of parabolas to note is their symmetry. The vertex acts as the line of symmetry, where both sides of the graph mirror each other. The point \((0, 16)\) serves as the peak for this parabola, since here, the parabola dips downward as \(x\) increases from zero.

• For \(f(x) = 16 - x^2\):
  • The vertex is at \((0, 16)\).
  • It intersects the x-axis at \((4, 0)\).

Domain Restrictions

Domain restrictions play a crucial role in determining which parts of a graph are valid.
For \(f(x) = 16 - x^2\), the specified domain is \(x \geq 0\), meaning we only consider the right side of this parabola.
This truncation at \(x = 0\) changes the graph from a full, symmetrical parabola to a half-parabola, beginning at its vertex.

• This dictates that the range for the original function is from the vertex down to the axis, specifically \(16 \geq y \geq 0\).
• When considering the inverse function, these domain restrictions influence its range to \(0 \leq x \leq 16\).

Grasping the constraints imposed by domain restrictions allows for accurate graph representations.