Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

Identify the critical points and find the maximum value and minimum value on the given interval. $$ h(x)=e^{-x^{2}} ; I=[-1,3] $$

Step-by-Step Solution

Answer

Maximum at \( x = 0 \), minimum at \( x = 3 \).

Step 1: Differentiate the Function

To find the maximum and minimum values, we need to find the critical points. First, differentiate the function. Given the function is \( h(x) = e^{-x^2} \), use the chain rule to differentiate: \[h'(x) = \frac{d}{dx} e^{-x^2} = e^{-x^2} \cdot \frac{d}{dx}(-x^2) = e^{-x^2} \cdot (-2x) = -2xe^{-x^2}. \]

Step 2: Find Critical Points

Critical points occur where the derivative is zero or undefined. Set the derivative equal to zero and solve for \( x \): \[-2xe^{-x^2} = 0.\] This equation is zero when \( x = 0 \) because \( e^{-x^2} \) is never zero. Thus, the critical point is at \( x = 0 \).

Step 3: Evaluate the Function at Critical Points and Endpoints

To find the maximum and minimum values on \( I = [-1, 3] \), evaluate \( h(x) \) at the critical point and at the endpoints of the interval.- Calculate \( h(-1) = e^{-(-1)^2} = e^{-1} \).- Calculate \( h(0) = e^{-0^2} = e^{0} = 1 \).- Calculate \( h(3) = e^{-(3)^2} = e^{-9} \).

Step 4: Compare the Function Values

Compare the function values at the critical point and the endpoints:- \( h(-1) = e^{-1} \approx 0.3679 \)- \( h(0) = 1 \)- \( h(3) = e^{-9} \approx 0.000123 \)The maximum value is 1 at \( x = 0 \), and the minimum value is \( e^{-9} \approx 0.000123 \) at \( x = 3 \).

Key Concepts

Calculus

Calculus is a branch of mathematics that studies how things change. It focuses on rates of change and how quantities are accumulated. It's fundamental in fields like physics, engineering, and economics.
Calculus is primarily divided into two branches: differential calculus and integral calculus.

• Differential Calculus: This deals with the concept of a derivative. It helps understand how a function changes at any given point, describing the rate of change at that point.
• Integral Calculus: This involves integrals and accumulation of quantities, identifying areas under curves.

In the context of finding critical points and determining maximum and minimum values, we rely on differential calculus.
Through calculus, we get a deeper insight into the behavior of functions, and this allows us to solve complex problems involving maxima and minima efficiently.

Function Differentiation

Differentiation is a critical concept in calculus that measures how a function changes as its input changes. To differentiate a function means to find its derivative, which represents the slope of the function at any point.When we talk about differentiating a function, we typically use rules such as:

• Power Rule: For a function of the form \( f(x) = x^n \), the derivative is \( f'(x) = nx^{n-1} \).
• Chain Rule: Used when differentiating a composite function, allowing you to "chain" together derivatives of individual functions.

For the exercise, we used the chain rule to differentiate the function \( h(x) = e^{-x^2} \), which resulted in the derivative \( h'(x) = -2xe^{-x^2} \).
Finding this derivative is the first step in identifying critical points where the function might reach a maximum or minimum value.

Maximum Value

The maximum value of a function is the greatest value that the function attains. In the context of calculus, we're often interested in finding local maxima, which are the highest points in a particular interval, or absolute maxima, which are the highest points in the entire domain of the function.To find the maximum value, follow these steps:

• Differentiate the function: Determine where the slope of the function is zero by finding where its derivative equals zero.
• Find critical points: These are potential points for maxima. They occur where the derivative is zero or undefined.
• Evaluate function at endpoints and critical points: In closed intervals, also check the function's values at the endpoints.

In our exercise, we evaluated \( h(x) \) at the critical point \( x = 0 \) and the endpoints \( x = -1 \) and \( x = 3 \).
We concluded the maximum value is 1, occurring at \( x = 0 \).

Minimum Value

The minimum value of a function is the lowest value that the function attains. Like finding maximum values, discovering minimum points involves examining where a function’s slope is zero or where the derivative is undefined.The process is fairly similar to finding the maximum value:

• Find the derivative of the function: This helps locate critical points by setting the derivative equal to zero or noting where it is undefined.
• Identify critical points and endpoints: Calculate function values at these points to evaluate minima.
• Compare function values: By comparing these values, determine the minimum.

In the specific exercise given, you evaluated \( h(x) \) at \( x = 3 \), where the value was approximately 0.000123.
This point, the endpoint of the interval, was identified as having the minimum value for the function.