Problem 20 - Calculus with Concepts in Calculus

Question

Find all extreme values (if any) of the given function on the given interval. Determine at which numbers in the interval these values occur. $$ g(x)=1 / x ;(0,3] $$

Step-by-Step Solution

Answer

The minimum value is at \( x = 3 \) with \( g(3) = \frac{1}{3} \); there is no maximum.

Step 1: Understand the Function

The given function is \( g(x) = \frac{1}{x} \), and we're considering the interval \((0, 3]\). We need to understand that as \(x\) approaches 0, \(g(x)\) approaches infinity, and as \(x\) increases toward 3, \(g(x)\) gets smaller.

Step 2: Find the Derivative

To find critical points, we calculate the derivative \( g'(x) \). Using the power rule, the derivative of \( g(x) = x^{-1} \) is: \[ g'(x) = -1 x^{-2} = -\frac{1}{x^2}. \]

Step 3: Determine Critical Points

Critical points occur where \( g'(x) = 0 \) or where \( g'(x) \) is undefined. Since \( -\frac{1}{x^2} \) never equals zero, there are no critical points where the derivative equals zero. However, \( g'(x) \) is undefined at \( x = 0 \). As this is not within the open interval, it does not affect the extreme values on the open interval (0,3].

Step 4: Evaluate the Function at Endpoints

To find extreme values, evaluate \( g(x) \) at the endpoints of the interval (although strictly speaking, \(x = 0\) is not included). Calculate: 1. \( g(3) = \frac{1}{3} \). Since the left-end of the interval \( x \to 0^+ \), \( g(x) \to +\infty \).

Step 5: Determine the Extreme Values

At \( x = 3 \), \( g(x) = \frac{1}{3} \) which is a local minimum, and as \( x \to 0^+ \), \( g(x) \to +\infty \), indicating there is no attainable maximum within \((0, 3]\).

Key Concepts

Interval Notation

Interval Notation is a way of writing subsets of the real number line. It is often used in calculus to specify the domain (or part of the domain) over which a function is analyzed. In the problem we have

1. \((0, 3]\) which denotes all numbers greater than 0 but less than or equal to 3.
   • The parenthesis \((\) indicates that 0 is not included in the interval, often referred to as 'open' at that endpoint.
   • The bracket \(\big[\big]\) shows that 3 is included, making it 'closed' at that endpoint.

This combination allows us to consider values near zero but does not allow zero itself, which is important when analyzing functions like \(g(x) = \frac{1}{x}\). It's because the function becomes undefined at zero.
Recognizing how intervals are expressed makes working with function limits and endpoints easier, especially when seeking extreme values.

Function Derivatives

In calculus, derivatives are crucial as they represent the rate of change of a function. For a function \(g(x) = \frac{1}{x}\), the derivative helps us understand how \(g(x)\) changes as \(x\) changes. To find the derivative of \(g(x)\) we make a simple conversion using the power rule.

• Write \(g(x)\) as \(x^{-1}\).
• Apply the power rule: Bring down the exponent and subtract one from the exponent. This gives \(g'(x) = -1 \cdot x^{-2} = -\frac{1}{x^2}\).

The key here is that this derivative shows us that \(g(x)\) decreases as \(x\) increases because the derivative is always negative in the interval. By understanding the derivative, we get insights into the behavior of \(g(x)\) and can determine critical points, leading us to find the function's extreme values.

Critical Points

Critical points occur where a function's derivative is zero or undefined. This often reveals potential maximum or minimum points of the function.
For \(g(x) = \frac{1}{x}\), we analyze its derivative \(g'(x) = -\frac{1}{x^2}\).

• This derivative is never zero, indicating no points where the function stops changing.
• However, it's undefined at \(x = 0\), a boundary of our interval.

Even though the derivative presents undefined behavior at zero, it's important to realize that zero isn't included within our interval \((0, 3]\). Hence, no critical points are found within the open portion of the interval.

Local Minimum

A local minimum is a point where a function's value is lower than all surrounding points. Even within a specific interval, identifying such points can be exceedingly helpful.
In our function \(g(x) = \frac{1}{x}\) over the interval \((0, 3]\), we observe:

• As \(x\) approaches zero, \(g(x)\) moves towards infinity, indicating no maximum.
• Evaluating \(g(x)\) at the endpoint \(x = 3\) gives us \(g(3) = \frac{1}{3}\).

This value is smaller than any other value of \(g(x)\) within the interval because \(g(x)\) decreases as \(x\) moves from 0 towards 3. Hence, \(x = 3\) is a local minimum within the interval. Understanding this concept highlights the importance of evaluating a function's endpoint to determine its extreme values in specified domains.