Problem 54 - 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. $$f(x)=\sqrt{x^{2}-2 x-3}, \quad 3 \leq x<\infty$$

Step-by-Step Solution

Answer

There are no local extreme values in the domain, and no absolute extreme values since the function increases beyond 3.

Step 1: Simplify the Function

We simplify the function by identifying the expression under the square root. The function is given as \( f(x) = \sqrt{x^2 - 2x - 3} \). First, rewrite the expression under the square root. It can be expressed as \( (x - 3)(x + 1) \) by factoring. Therefore, \( f(x) = \sqrt{(x - 3)(x + 1)} \).

Step 2: Determine the Domain

Describe the domain given in the problem: \( 3 \leq x < \infty \). This limits our consideration to values of \( x \) starting from 3 onwards.

Step 3: Find the Derivative

To find local extrema, we first find the derivative of \( f(x) \). Using the chain rule, the derivative \( f'(x) \) is computed as follows: \[ f'(x) = \frac{1}{2\sqrt{x^2 - 2x - 3}} \cdot (2x - 2) \] Which simplifies to: \[ f'(x) = \frac{x - 1}{\sqrt{(x - 3)(x + 1)}} \].

Step 4: Find Critical Points

Set \( f'(x) = 0 \) to find the critical points. Solving: \[ \frac{x - 1}{\sqrt{(x - 3)(x + 1)}} = 0 \] leads to \( x - 1 = 0 \) implying \( x = 1 \). However, \( x = 1 \) is not in the domain \( 3 \leq x < \infty \), thus no critical points for local extrema are within the domain.

Step 5: Assess for Absolute Extremes

Evaluate \( f(x) \) at the endpoints of the domain. Since 3 is the only meaningful boundary point, substitute into the function: \[ f(3) = \sqrt{3^2 - 2 \times 3 - 3} = \sqrt{0} = 0 \]. As the function continues to increase as \( x \) increases within the domain, \( f(x) \) does not reach a higher value than at any endpoint; hence, function values greater than zero indicate no other extremes are absolute.

Step 6: Graph Support

Plot the function on a graph for the domain starting at 3. The graph should show that \( f(x) \) begins at 0 for \( x = 3 \) and increasing as \( x \) grows. The graph confirms the lack of peaks or valleys within the domain.

Key Concepts

Local Extrema

Local extrema refer to the points on a function where it reaches either a local maximum or local minimum. These are locations on the curve where the function changes direction.

• Local Maximum: This is where a function value is higher than all nearby points.
• Local Minimum: This is where a function value is lower than all nearby points.

To find local extrema, we typically find the derivative of the function and set it to zero to identify critical points. However, it's crucial to remember these points must also lie within the function's domain.
In our given function, we found its critical point at \(x = 1\) which is not within the domain \(3 \leq x < \infty\). Therefore, there are no local extrema within this domain.

Absolute Extrema

Absolute extrema are the highest and lowest points of a function on its entire domain. An absolute maximum is the highest point, while an absolute minimum is the lowest.
To identify absolute extrema of a function, we evaluate it at any critical points and also at the endpoints of the specified domain.
For the function \(f(x) = \sqrt{(x-3)(x+1)}\), the domain starts at \(x = 3\). Evaluating at this point gives \(f(3) = 0\). Since the function is increasing beyond \(x = 3\), \(f(x)\) has no upper bound in the domain, exhibiting no absolute maximum.
Thus, the absolute minimum within the given domain is \(f(3) = 0\).

Derivative

A derivative, denoted as \(f'(x)\), is the rate at which a function changes at any given point. It provides critical information regarding the slope of the function.
Knowing the derivative helps identify potential local extrema. When \(f'(x) = 0\), the function may have a local maximum, minimum, or an inflection point.
For our function \(f(x) = \sqrt{(x - 3)(x + 1)}\), the derivative is \(f'(x) = \frac{x - 1}{\sqrt{(x-3)(x+1)}}\). Setting \(f'(x) = 0\) gives critical points where potential extrema might be found, though these points must lie within the allowable domain.
Through this process, although \(x = 1\) appears as a critical point, it is not within our domain and therefore does not contribute to finding a local extremum.

Graphing Functions

Graphing functions visually represents a function's behavior over a defined domain. It helps to see where a function increases, decreases, and holds constant. Graphs also reveal the location and nature of extrema.
Using a graphing calculator or software, we plotted \(f(x)\) for \(x \geq 3\). The graph shows the function starts from zero and rises as \(x\) increases, confirming analyses made via calculus.
The absence of peaks and valleys in the graph supports the conclusion that there are no local extrema. Moreover, seeing \(f(x)\) rise validate no absolute maximum exists, and also visually confirming the absolute minimum at \(x = 3\).
Graphing aids in visualizing function characteristics, solidifying understanding beyond algebraic calculations.