Problem 90 - A Graphical Approach to Precalculus with Limits

Question

the domains were determined in Exercises \(73-80 .\) Use a graph to (a) find the range,(b) give the largest open interval over which the finction is increasing. (c) give the largest open interval over which the finction is decreasing, and (d) solve the equation \(f(x)=0\) by observing the graph. $$f(x)=\sqrt{81-x^{2}}$$

Step-by-Step Solution

Answer

(a) Range: [0, 9]; (b) Increasing: (-9, 0); (c) Decreasing: (0, 9); (d) Solutions: x = -9, 9.

Step 1: Understand the Function

The given function is \( f(x) = \sqrt{81 - x^2} \), which is the equation of a semicircle. The graph represents the upper half of a circle centered at the origin with radius 9.

Step 2: Determine the Domain and Range

The domain of the function \( f(x) = \sqrt{81 - x^2} \) is \(-9 \leq x \leq 9\) since the expression under the square root must be non-negative. For the range, as \(x\) varies from \(-9\) to \(9\), \(f(x)\) takes values from \(0\) to \(9\). Hence, the range is \([0, 9]\).

Step 3: Identify Increasing Interval

The function is increasing where the derivative \( f'(x) \) is positive. Since the semicircle opens downward, \( f(x) \) increases as \( x \) moves from \(-9\) to \(0\). Thus, the largest open interval over which the function is increasing is \((-9, 0)\).

Step 4: Identify Decreasing Interval

The function is decreasing where the derivative \( f'(x) \) is negative. For the function \( f(x) = \sqrt{81 - x^2} \), after \(x = 0\), it decreases as \(x\) moves towards \(9\). The largest open interval where the function is decreasing is \((0, 9)\).

Step 5: Solve the Equation \(f(x) = 0\)

To solve \( f(x) = 0 \), we need \( \sqrt{81 - x^2} = 0 \). This simplifies to \( 81 - x^2 = 0 \) or \( x^2 = 81 \). Solving for \( x \), we get \( x = \pm 9 \). Hence, the solutions to \(f(x) = 0\) are \( x = -9 \) and \( x = 9 \).

Key Concepts

Domain of a Function

When we talk about the domain of a function, we mean all possible input values (\(x\)-values) that can be used in the function without causing any mathematical errors. For the function \(f(x) = \sqrt{81 - x^2}\), since it's a square root function, what's under the square root must remain non-negative.
This is crucial because the square root of a negative number is not defined in the set of real numbers. To ensure mathematical correctness:

• We set \(81 - x^2 \geq 0\). This inequality has to hold true for all input \(x\)
• Solving \(81 - x^2 \geq 0\) gives us \(-9 \leq x \leq 9\)

This solution tells us that the domain of \(f(x)\) is all real numbers between -9 and 9, inclusive. It ensures that our function outputs only valid numbers, keeping it defined and functional over the specified range.

Increasing and Decreasing Intervals

For any function, determining where it is increasing or decreasing is essential for understanding its overall behavior. When the function rises as \(x\) increases, we say it is increasing. Conversely, it is decreasing if the function descends as \(x\) moves in the positive direction. Let's look at \(f(x) = \sqrt{81 - x^2}\) to understand its behavior:

• Increasing Intervals: The function increases when the graph goes upwards as we progress along the \(x\)-axis. Due to the semicircular shape, \(f(x)\) increases as \(x\) moves from -9 to 0.The largest open interval where the function increases is \((-9, 0)\)
• Decreasing Intervals: A function is decreasing when its graph goes downward. For the semicircle of \(f(x)\), it starts to decrease after \(x=0\) and continues until \(x=9\).The largest interval over which the function is decreasing is \((0, 9)\)

Recognizing these intervals is vital for graphically analyzing and interpreting functions.

Solving Equations Graphically

Solving equations graphically involves finding the points where the function intersects key points on the graph, usually the \(x\)-axis. For the equation \(f(x) = 0\), we need to identify where the curve touches or crosses the \(x\)-axis:

• The function \(f(x) = \sqrt{81 - x^2}\) has zeros at the points where the semicircle meets the \(x\)-axis.
• Setting \(f(x) = 0\) simplifies to solving \(\sqrt{81 - x^2} = 0\), which leads to \(81 - x^2 = 0\).
• Solving this gives \(x^2 = 81\), or \(x = \pm 9\). Hence, the points \(x = -9\) and \(x = 9\) satisfy the equation.

These solutions indicate where the function intersects the \(x\)-axis graphically. These graphical solutions provide a visual representation of where the output equals zero, helping us see the function's behavior clearly.