Problem 16 - Calculus of a Single Variable

Question

In Exercises 5–24, analyze and sketch a graph of the function. Label any intercepts, relative extrema, points of inflection, and asymptotes. Use a graphing utility to verify your results. $$ g(x)=x \sqrt{9-x^{2}} $$

Step-by-Step Solution

Answer

The function \( g(x)=x \sqrt{9-x^{2}} \) has a domain of \( -3 \leq x \leq 3 \), x-intercepts at \( x = -3 \) and \( x = 3 \), y-intercept at \( y = 0 \), and a relative maximum at \( x=0 \). There are no points of inflection and asymptotes in the domain. These results can be verified using a graphing utility.

Step 1: Identify the domain

Firstly, calculate the domain of the function. The expression under the square root \( (9-x^{2}) \) cannot be negative, so the domain of \( g(x) \) is \( -3 \leq x \leq 3 \).

Step 2: Calculate the intercepts

The x-intercepts are found by setting \( g(x) = 0 \). Even though we have a multiplication, note that if either \( x \) or \( 9-x^{2} \) equals 0, the product equals 0, Thus, we have two x-intercepts at \( x = -3 \) and \( x = 3 \). To find the y-intercept(s), we know that \( g(0) = 0 \) therefore the y-intercept is at \( y = 0 \).

Step 3: Compute the extrema

To find out if there are any relative max or min points (extrema), we need to take the derivative of the function and set it to 0. Using the product rule, we get: \( g'(x) = \sqrt{9 - x^{2}} + x(-x/\sqrt{9 - x^{2}}) \). Solving this equation we get \( g'(x)=0 \) when \( x=0 \). Thus, the function has an extremum at \( x=0 \). In this case, it's a relative maximum.

Step 4: Identify points of inflection and asymptotes

The points of inflection are found by getting the second derivative and setting it to 0. However, for this particular function, there are no points of inflection and asymptotes within its domain.

Step 5: Sketch and verify graph

By combining all the above information, we can sketch the graph of the function. Always refer to a graphing utility to double-check the result and make sure to label all important points (x and y intercepts, extrema), and make clear the domain and range of the function.

Key Concepts

Function Intercepts

Intercepts are the points where the graph of a function crosses the axes. Understanding intercepts helps us understand the behavior of the function as it interacts with the coordinate plane.

• X-Intercept(s): The x-intercept(s) occur where the function equals zero. For the function \( g(x) = x \sqrt{9-x^{2}} \), we set it to zero. Thus, we evaluate \( x \) and the expression \( 9 - x^{2} \). If either is zero, \( g(x) \) equals zero. This gives x-intercepts at \( x = -3 \) and \( x = 3 \).
• Y-Intercept: The y-intercept occurs where \( x = 0 \). Plugging \( x = 0 \) into the function gives \( g(0) = 0 \). Thus, the y-intercept is at \( y = 0 \).

Finding these intercepts is crucial for graph analysis, providing key anchor points where the graph intersects the x-axis or y-axis.

Domain and Range

The domain of a function refers to the set of all possible input values (x-values), while the range refers to the set of all possible output values (y-values).

To find the domain of the function \( g(x) = x \sqrt{9-x^{2}} \), observe the square root's expression, \( 9-x^{2} \), which must be non-negative. To ensure non-negativity:

• Set \( 9-x^{2} \geq 0 \), resulting in \( -3 \leq x \leq 3 \). This is the domain of the function.

The range is determined by evaluating the output values within the domain:

• Since \( x \) and \( \sqrt{9-x^{2}} \) both yield 0 at the endpoints \( x = -3, 3 \), and \( g(x) \) reaches a maximum value at \( x = 0 \) with \( y = 0 \), this suggests the function ranges over \(-\infty, \infty\), as the function slopes from zero up to its maximum point.

Determining both domain and range helps understand which values are valid for the function and how those values behave.

Relative Extrema

Relative extrema refer to the highest or lowest points in a specified interval of the function. These extrema can be either relative minimums or maximums within that interval.

To find relative extrema for \( g(x) = x \sqrt{9-x^{2}} \), we need to find where the derivative \( g'(x) \) equals zero. Calculating the derivative using product rule:

\[ g'(x) = \sqrt{9 - x^{2}} + x\left(-\frac{x}{\sqrt{9 - x^{2}}}\right) \]

Set \( g'(x) = 0 \):

• Solving, you find \( g'(x) = 0 \) at \( x = 0 \).

This indicates a relative extremum at \( x = 0 \). Since this also shows that it achieves its maximum here, it's identified as a relative maximum. Understanding relative extrema gives insight into the peaks and valleys within the function's graph, illustrating where a function increases, decreases, or remains constant.

Derivative Calculation

Derivatives are essential for understanding a function's rate of change. They tell us how the function's output value changes with respect to changes in the input value. Calculating derivatives can help find relative extrema and other critical points in a function's graph.

To find the derivative of \( g(x) = x \sqrt{9-x^{2}} \), you use the product rule. The product rule states:

• If \( f(x) = u(x) \cdot v(x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \).

Applying this to \( g(x) \):

\[ g'(x) = \sqrt{9-x^{2}} + x \left(-\frac{x}{\sqrt{9-x^{2}}}\right) \]

Here:

• \( u = x \)
• \( v = \sqrt{9-x^{2}} \)
• \( u' = 1 \)
• \( v' = \frac{-x}{\sqrt{9-x^{2}}} \)

This derivative calculation helps find where the function's slope is zero, indicating potential relative minima or maxima, and provides a deeper analysis of the function's behavior across its domain.