Problem 40 - University Calculus: Early Transcendentals

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. $$g(x)=e^{-x^{2}}, \quad-2 \leq x \leq 1$$

Step-by-Step Solution

Answer

The absolute maximum is 1 at \( x = 0 \), and the absolute minimum is \( e^{-4} \) at \( x = -2 \).

Step 1: Identify Critical Points

We find the derivative of the given function to locate critical points. The function is \( g(x) = e^{-x^2} \). Using the chain rule, the derivative is \( g'(x) = -2xe^{-x^2} \). We set \( g'(x) = 0 \) to find critical points: \(-2xe^{-x^2} = 0 \). This implies \( x = 0 \). This is the only critical point inside the interval \([-2, 1]\).

Step 2: Evaluate at Critical Points and Endpoints

We evaluate the function at the critical point and the endpoints of the interval \([-2, 1]\).- At \( x = -2 \), \( g(-2) = e^{-(-2)^2} = e^{-4} \).- At \( x = 0 \), \( g(0) = e^{0} = 1 \).- At \( x = 1 \), \( g(1) = e^{-1^2} = e^{-1} \).

Step 3: Determine the Absolute Extrema

By comparing the values, we identify the absolute maximum and minimum. - The absolute maximum value is \( g(0) = 1 \) at \( x = 0 \).- The absolute minimum value is \( g(-2) = e^{-4} \) at \( x = -2 \).

Step 4: Graph the Function

On the graph of \( g(x) = e^{-x^2} \), plot the critical point and the endpoints with their coordinates. - Mark the point (0, 1) for the maximum.- Mark the point (-2, \( e^{-4} \)) for the minimum.- Plot the point (1, \( e^{-1} \)) to confirm the function's decreasing behavior over the right side of the interval.

Key Concepts

Critical Points

Critical points are where the derivative of a function equals zero or is undefined. These points are essential in finding the function's extrema, as maxima or minima may occur there. For our function, we began by determining the derivative of the function \( g(x) = e^{-x^2} \). Using the chain rule, the derivative is computed as \( g'(x) = -2xe^{-x^2} \). We then set the derivative equal to zero, \( -2xe^{-x^2} = 0 \), to find the critical points. Simplifying this equation, we get \( x = 0 \), which is our critical point. It’s vital to observe that this point is within our considered interval of \([-2, 1]\). Critical points help us discover where potential extrema may appear.

Derivative

In calculus, the derivative represents the rate of change or the slope of the function at any given point. Understanding derivatives is crucial for finding critical points and analyzing the behavior of a function. In the exercise, we calculated the derivative of \( g(x) = e^{-x^2} \) as \( g'(x) = -2xe^{-x^2} \). By using the chain rule, we successfully found how the function's output changes as \( x \) changes. The derivative provides insights into where the function increases or decreases. Such information is key to locating all turning points and testing the conditions for maximum or minimum values.

Interval Evaluation

Once we have determined the critical points, the next step is interval evaluation. This step involves testing the function at all critical points and the boundaries of the interval. For the interval \([-2, 1]\), we computed the function's values at the points \(-2, 0,\) and \(1\).

• At \(x = -2\), the function \(g(x)\) evaluates to \( e^{-4} \).
• At \(x = 0\), we find \(g(0) = 1\).
• At \(x = 1\), we compute \(g(1) = e^{-1}\).

By comparing these values, we identified the absolute maximum as \(g(0) = 1\) and the absolute minimum as \(g(-2) = e^{-4}\). Evaluating at these points ensures that we capture all potential extrema within the interval.

Graphing Functions

Graphing functions is a vital way to visualize their behavior over a given interval. Plotting can reveal critical points and the nature of the extrema. For \( g(x) = e^{-x^2} \) on \([-2, 1]\), we graphically verified our calculated extrema. The function's smooth, bell-shaped curve highlights its decrease as \( x \) moves away from zero toward \(-2\) and \(1\).

• The absolute maximum at \(x = 0\), with coordinates \((0, 1)\), is marked clearly.
• The minimum at \(x = -2\), corresponding to \(( -2, e^{-4})\), is shown on the left end.
• The graph at \(( 1, e^{-1})\) confirms the diminished value as \(x\) approaches 1.

By observing these points on the graph, students can confirm the smooth transition and the extremas of \( g(x) = e^{-x^2} \). Graphs are an excellent tool to intuitively understand the function's behavior and to visually confirm computed values.