Problem 106 - A Graphical Approach to College Algebra

Question

Suppose \(f(r)\) is the volume (in cubic inches) of a sphere of radius \(r\) inches. What does \(f^{-1}(5)\) represent?

Step-by-Step Solution

Answer

\(f^{-1}(5)\) is the radius of a sphere with volume 5 cubic inches.

Step 1: Understanding the Function

The function \( f(r) \) gives the volume of a sphere with radius \( r \), which is mathematically expressed as \( f(r) = \frac{4}{3}\pi r^3 \). The function maps radius to volume.

Step 2: Understanding the Inverse Function

The inverse function \( f^{-1}(x) \) does the opposite of \( f(x) \); it takes a volume and returns the radius. Therefore, \( f^{-1}(5) \) takes the volume \( 5 \) cubic inches and returns the radius of the sphere with this volume.

Step 3: Expression for Radius from Volume

To express \( r \) in terms of \( V \), we rearrange the volume formula: \( V = \frac{4}{3}\pi r^3 \). Solving for \( r \), we have \( r = \left(\frac{3V}{4\pi}\right)^{1/3} \).

Step 4: Substitute and Solve

Substitute \( V = 5 \) into the radius formula: \( r = \left(\frac{3 \times 5}{4\pi}\right)^{1/3} \). Simplify the expression to find the numerical value of \( r \).

Key Concepts

Volume of a Sphere

The volume of a sphere is a mathematical concept that describes how much space a sphere occupies. This measurement is expressed in cubic units. To calculate the volume of a sphere, we use the formula:\[V = \frac{4}{3}\pi r^3\]Here,

• \( V \) represents the volume.
• \( r \) is the radius of the sphere.
• \( \pi \) is a constant that is approximately equal to 3.14159.

The formula tells us that the volume increases with the cube of the radius. This means even small increases in the radius will lead to much larger increases in volume.
So, when considering spheres, the radius is crucial in determining how much space a sphere will occupy.

Radius and Volume Relationship

The relationship between the radius and the volume of a sphere is central to understanding spheres. The formula \( V = \frac{4}{3}\pi r^3 \) clearly shows the dependency of volume on the cube of the radius, \( r \).

• As the radius increases, the sphere's volume increases rapidly.
• Doubling the radius results in an eightfold increase in volume, demonstrating how changes in radius can significantly impact the volume of a sphere.

Understanding this relationship is important in many real-world applications. For instance, when designing objects that should contain or displace a certain amount of liquid or air efficiently, the radius is a factor that heavily influences the capacity or displacement.

Rearranging Formulas

Rearranging formulas is a key skill in mathematics and science. It involves changing the order of a formula to express a different variable as the subject of the equation. In the context of a sphere's volume, rearranging the formula allows us to express the radius in terms of the volume.Given the formula for the volume of a sphere, \( V = \frac{4}{3}\pi r^3 \), to solve for \( r \), follow these steps:1. Multiply both sides by the reciprocal of \( \frac{4}{3} \) to isolate \( r^3 \): \[ r^3 = \frac{3V}{4\pi} \]2. Take the cube root of both sides to solve for \( r \): \[ r = \left(\frac{3V}{4\pi}\right)^{1/3} \]Rearranging allows us to compute the necessary radius for any given volume, which is particularly useful for finding the dimensions of objects when only the volume is known. This kind of manipulation is handy in fields such as engineering, architecture, and any area involving physical space calculations.