Problem 53 - University Calculus: Early Transcendentals

Question

a. Identify the function's local extreme values in the given domain, and say where they occur. b. Which of the extreme values, if any, are absolute? c. Support your findings with a graphing calculator or computer grapher. $$f(x)=\sqrt{25-x^{2}}, \quad-5 \leq x \leq 5$$

Step-by-Step Solution

Answer

The local and absolute maximum is 5 at \(x = 0\), and the absolute minimum is 0 at \(x = -5\) and \(x = 5\).

Step 1: Understanding the Function

The function given is \( f(x) = \sqrt{25 - x^2} \). This is a semicircle with a radius of 5 units and is centered at the origin. The domain given, \(-5 \leq x \leq 5\), represents the entire horizontal range of this semicircle.

Step 2: Determine Critical Points

To find critical points, we need the derivative of \( f(x) \): the derivative is \( f'(x) = -\frac{x}{\sqrt{25-x^2}} \). Setting this equal to zero, we solve for \( x \): \( -\frac{x}{\sqrt{25-x^2}} = 0 \), leading to \( x = 0 \). Thus, \( x = 0 \) is a critical point.

Step 3: Evaluate the Function at Critical and Endpoint Values

Evaluate \( f(x) \) at the critical point and endpoints: \( f(-5) = \sqrt{25 - (-5)^2} = 0 f(0) = \sqrt{25 - 0^2} = 5 f(5) = \sqrt{25 - 5^2} = 0\). These results tell us the function values at critical points and endpoints.

Step 4: Determine Local Extreme Values

In the interval \([-5, 5]\), \( f(x) \) has local maximum at \( x = 0 \) because \( f(0) = 5 \) is greater than \( f(-5) \) and \( f(5) \), which are both zero. Hence, the local maximum value is 5 and occurs at \( x = 0 \).

Step 5: Identify Absolute Extreme Values

Since the highest function value \( f(0) = 5 \) is within the bounds and there are no other values greater than 5, this is also the absolute maximum. The values at endpoints \( f(-5) = 0 \) and \( f(5) = 0 \) are the smallest, making zero the absolute minimum.

Step 6: Use Graphing Technology for Verification

Graph \( f(x) = \sqrt{25 - x^2} \) using a graphing calculator or software. Confirm visually that the semicircle reaches its peak at \( x = 0 \) with a height of 5 and descends back to 0 at the endpoints \( x = -5 \) and \( x = 5 \). This verifies our calculated extreme values.

Key Concepts

Critical Points

Critical points are where a function's derivative is either zero or undefined. We find these points because they can indicate potential local extreme values. For the function given,

• Start by finding the derivative: for \( f(x) = \sqrt{25 - x^2} \), the derivative is calculated as \( f'(x) = -\frac{x}{\sqrt{25-x^2}} \).
• Find where this derivative equals zero: solve \(-\frac{x}{\sqrt{25-x^2}} = 0\), which simplifies to \(x = 0\).
• No points where the derivative is undefined because the expression under the square root is positive in the given domain \([-5, 5]\).

Therefore, the critical point for this function is at \(x = 0\). Checking these points helps determine where changes in direction (like peaks or troughs) occur.

Absolute Extreme Values

Absolute extreme values refer to the absolute highest or lowest values that a function achieves on a given interval. These values can occur at critical points or endpoints of the interval. To determine them for our function, we evaluated:

• The function at the critical point \(x = 0\) yielding \(f(0) = 5\).
• The function at the endpoints \(x = -5\) and \(x = 5\) giving \(f(-5) = 0\) and \(f(5) = 0\).

Comparing these, \(f(0) = 5\) is the largest, making it the absolute maximum. Likewise, \(f(-5)\) and \(f(5)\) are equal to zero, representing the absolute minimum values for this interval. Absolute extremes help understand the overall behavior of the function over a specified range.

Derivative

The derivative of a function gives us the slope or rate of change of the function at any given point. It's crucial in finding critical points and, ultimately, analyzing extreme values. For the function \(f(x) = \sqrt{25 - x^2}\), the derivative is determined as:

• Using implicit differentiation on the expression under the square root: \(f'(x) = -\frac{x}{\sqrt{25-x^2}}\).
• The negative sign indicates the slope direction is downward as \(x\) increases or decreases from zero within the domain.

This derivative expression helps locate where the slope is zero, leading to critical points and understanding local or absolute extrema. Understanding how a derivative works is essential for tackling calculus problems and visualizing how functions behave graphically.