Problem 11 - Single Variable Calculus

Question

In Exercises \(7-24\), sketch the graph of the function and find its absolute maximum and absolute minimum values, if any. $$ g(x)=x^{2}+1 \text { on }(0, \infty) $$

Step-by-Step Solution

Answer

The function \(g(x)=x^2+1\) has an absolute minimum value of 1, which occurs at the open endpoint at \(x = 0\), and no absolute maximum value.

Step 1: Find the derivative of the function

First, we need to find the derivative of the function \(g(x)\). This will help us identify the critical points where the function has a maximum or minimum value. $$ g'(x) = \frac{d}{dx} (x^2 + 1) = 2x $$

Step 2: Find the critical points

Now find the critical points by setting the derivative equal to zero: $$ 2x = 0 \\ x = 0 $$ So, there is only one critical point at \(x = 0\).

Step 3: Analyze the endpoints

We must also consider the endpoints of the interval \((0, \infty)\) to determine the absolute minimum and maximum values. Since the domain only includes values larger than 0, our only endpoint to consider is at \(x = 0\). However, this point is not included in the interval, so we only have the critical point \(x = 0\) to analyze.

Step 4: Determine the absolute maximum and minimum

Now that we have found the only critical point, we will analyze the values of the function at this point: $$ g(0) = (0)^2 + 1 = 1 $$ Since there are no other critical points and the function increases as x goes to infinity, the absolute minimum value is 1, which occurs at the point \((0,1)\). Note that the absolute minimum occurs at the open endpoint, as the point \((0,1)\) is not included in the interval. As \(x\) approaches infinity, \(g(x)\) also approaches infinity, so there is no absolute maximum value. To summarize: - Absolute minimum value: 1 (occurs at the open endpoint at \(x = 0\)) - Absolute maximum value: None

Key Concepts

Derivative

Understanding the derivative is crucial in calculus problems, as it provides insights into the behavior of functions. The derivative of a function represents the rate at which the function's value is changing. In simpler terms, it tells us how steep a function is at any given point. For our function, \(g(x) = x^2 + 1\), the derivative is found by taking the derivative of each term:

• \( \frac{d}{dx}(x^2) = 2x \)
• \( \frac{d}{dx}(1) = 0 \)

Thus, the derivative of the entire function is \(g'(x) = 2x\).
This derivative function helps us identify where the slope of the original function is zero or undefined, which are potential points for maximum and minimum values.

Critical points

Critical points are values of \(x\) where the derivative of a function is zero or undefined. At these points, the function could have a local maximum, minimum, or a stationary point where the function levels out.
For the given exercise, we found the derivative \(g'(x) = 2x\). To find critical points, we set this derivative equal to zero:

• \(2x = 0\)
• \(x = 0\)

This result shows that \(x = 0\) is a critical point. However, remember that \(x = 0\) is not included in the open interval \((0, \infty)\), which means it is not considered in our evaluation of absolute maximum or minimum values. Despite not being part of the interval, it informs us about the behavior of the function around this point.

Absolute maximum and minimum

In calculus, the absolute maximum and minimum are the highest and lowest values that a function reaches over a given interval. Unlike local extremums that occur in the vicinity of critical points, absolute extremums take into account the entire interval and endpoints.
For \(g(x) = x^2 + 1\) on \((0, \infty)\), the challenge is evaluating endpoints since the interval is open at \(x = 0\) and extends to infinity.

• At \(x = 0\), \(g(0) = 1\). Since \(x = 0\) is not part of the open interval, we consider this point as an endpoint for analysis purposes only.
• As \(x\) tends towards infinity, the value of \(g(x)\) increases endlessly due to the \(x^2\) term. Thus, there is no absolute maximum, as the function doesn't reach a highest point.

Conclusively, the absolute minimum value of the function is 1 at the open end near \(x = 0\). On the other hand, there is no absolute maximum as the function only increases when moving further into the interval.