Problem 30 - Thomas Calculus in SI Units

Question

Find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates. $$g(x)=-\sqrt{5-x^{2}}, \quad-\sqrt{5} \leq x \leq 0$$

Step-by-Step Solution

Answer

Absolute maximum is 0 at \((-\sqrt{5}, 0)\); absolute minimum is \(-\sqrt{5}\) at \((0, -\sqrt{5})\).

Step 1: Find the Critical Points

To find the critical points of the function, we first need to calculate the derivative of the function. The function is given by \(g(x) = -\sqrt{5-x^2}\). Apply the chain rule to find the derivative: \(g'(x) = \frac{-1}{2\sqrt{5-x^2}} \cdot (-2x) = \frac{x}{\sqrt{5-x^2}}\). Set \(g'(x) = 0\) to find critical points. Solving \(\frac{x}{\sqrt{5-x^2}} = 0\) gives the critical point \(x = 0\).

Step 2: Evaluate the Function at Boundaries and Critical Points

Evaluate \(g(x)\) at the endpoints and the critical point. The endpoints of the interval are \(x = -\sqrt{5}\) and \(x = 0\), and the critical point is \(x = 0\).\\[g(-\sqrt{5}) = -\sqrt{5 - (-\sqrt{5})^2} = -\sqrt{5-5} = 0\] \\[g(0) = -\sqrt{5 - 0^2} = -\sqrt{5}\]

Step 3: Determine the Absolute Maximum and Minimum

From Step 2, we have the function values: \(g(-\sqrt{5}) = 0\) and \(g(0) = -\sqrt{5}\). The absolute maximum value is 0, and the absolute minimum value is \(-\sqrt{5}\). Thus, the absolute maximum occurs at \((-\sqrt{5}, 0)\) and the absolute minimum occurs at \((0, -\sqrt{5})\).

Step 4: Sketch the Graph

Graph the function \(g(x) = -\sqrt{5-x^2}\) over the interval \([-\sqrt{5}, 0]\). It's a portion of the upper half of a circle centered at the origin with radius \(\sqrt{5}\), flipped upside down. Mark the absolute maximum at \((-\sqrt{5}, 0)\) and the absolute minimum at \((0, -\sqrt{5})\). These points are where the absolute extrema occur.

Key Concepts

Critical Points

In calculus, critical points are essential in identifying places where the graph of a function might have a maximum, minimum, or change direction. To find them, you first need to take the derivative of the function. The derivative represents the rate of change, or the slope, of the function. Here, the function is given by \( g(x) = -\sqrt{5-x^2} \).

• Apply the chain rule to differentiate \( g(x) \).
• The derivative \( g'(x) = \frac{x}{\sqrt{5-x^2}} \).
• Set \( g'(x) = 0 \) to find the critical points, which gives \( x = 0 \).

The concept of critical points is vital because these points could indicate where the function takes on a local minimum or maximum value. However, in this exercise, with an endpoint interval provided, it's crucial to consider not just where the derivative equals zero, but also the endpoint values themselves.

Absolute Extrema

The absolute extrema of a function are the highest and lowest values that the function takes on a given interval. To find them, evaluate the function at its critical points and endpoints. In this context, we have a function, \( g(x) = -\sqrt{5-x^2} \), evaluated on the closed interval \([-\sqrt{5}, 0]\).

• Evaluate \( g(x) \) at \( x = -\sqrt{5} \) and \( x = 0 \).
• For \( x = -\sqrt{5} \), \( g(-\sqrt{5}) = 0 \).
• For \( x = 0 \), \( g(0) = -\sqrt{5} \).

From these calculations, it's clear that the absolute maximum value of the function on the interval is 0 at \( x = -\sqrt{5} \), and the absolute minimum value is \(-\sqrt{5}\) at \( x = 0 \). The absolute extrema are imperative as they provide the overall peak and lowest dip for the function along the specified interval.

Graphing Functions

Graphing a function allows us to visualize how it behaves over an interval, including where its maximum and minimum values occur. For our exercise, we're interested in graphing \( g(x) = -\sqrt{5-x^2} \) over the interval \([-\sqrt{5}, 0]\).

• This equation represents the upper half of a circle centered at the origin with a radius of \( \sqrt{5} \) but upside down due to the negative sign.
• As we graph, it's essential to plot points, especially the extrema, which are at \(( -\sqrt{5}, 0 )\) and \(( 0, -\sqrt{5} )\).
• Connect these points smoothly to show the curve over the specified interval.

The graph visually confirms that the absolute maximum occurs at \(( -\sqrt{5}, 0 )\) and the absolute minimum at \(( 0, -\sqrt{5} )\). Graphing helps cement our findings by providing a clear and intuitive image of where these critical values lie.