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Chapter 10: Problem 35

Find the absolute maximum value and the absolute minimum value, if any, of each function. $$ f(x)=e^{-x^{2}} \text { on }[-1,1] $$

Short Answer

Expert verified
The absolute maximum value of the function \(f(x) = e^{-x^2}\) on the interval \([-1, 1]\) is 1 at \(x = 0\), and the absolute minimum value is \(e^{-1}\) at \(x = -1\) and \(x = 1\).

Step by step solution

01

Determine the derivative of the function

To find critical points, we first need to find the derivative of the function \(f(x) = e^{-x^2}\). Using the chain rule, we get: $$ f'(x) = (e^{-x^2})' = -2xe^{-x^2} $$
02

Find critical points

Next, set the derivative equal to 0 and solve for x to find the critical points: $$ f'(x) = -2xe^{-x^2} = 0 $$ This equation is satisfied when \(x = 0\), so \(x = 0\) is our only critical point.
03

Evaluate the function at critical points and endpoints

Now we will evaluate \(f(x)\) at the critical point and endpoints of the interval \([-1, 1]\): $$ f(0) = e^{-0^2} = e^0=1 $$ $$ f(-1) = e^{-(-1)^2} = e^{-1} $$ $$ f(1) = e^{-1^2} = e^{-1} $$
04

Determine the absolute maximum and minimum values

By comparing these function values, we can determine the absolute maximum and minimum values of the function on \([-1, 1]\): - The maximum value occurs at \(x = 0\), with \(f(0) = 1\). - The minimum value occurs at \(x = -1\) and \(x = 1\), with \(f(-1) = f(1) = e^{-1}\). Therefore, the absolute maximum value of the function is 1 and the absolute minimum value is \(e^{-1}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
When working with functions in calculus, identifying critical points is essential. Critical points are where the derivative of a function equals zero or is undefined. At these points, the function may have a local maximum, a local minimum, or a saddle point. To find these, you first find the derivative of the function. For example, if you have a function like \(f(x) = e^{-x^2}\), you'd need to apply the Chain Rule to differentiate it, resulting in the derivative \(f'(x) = -2xe^{-x^2}\).
  • Setting this derivative equal to zero \(-2xe^{-x^2} = 0\), helps us find our critical points. For this function, it's clear that when \(x = 0\), the derivative equals zero, marking it as a critical point.
  • Critical points indicate places to check further as potential locations for local extremum values, which can help in determining the overall behavior and important characteristics of a function.
Chain Rule
The Chain Rule is a fundamental technique in calculus used for finding derivatives of composite functions. It enables you to differentiate two or more functions that are nested within each other. Imagine an onion: just like you'd peel layers off, using the Chain Rule helps you "peel off" layers of a function step by step.
  • In terms of formula, if you have a function \(f(g(x))\), its derivative is \(f'(g(x)) \cdot g'(x)\). This means you differentiate the outer function while keeping the inner function unchanged, then multiply by the derivative of the inner function.
  • For the function \(f(x) = e^{-x^2}\), using the Chain Rule gives the derivative result \(f'(x) = -2xe^{-x^2}\) by differentiating the outside function \(e^u\) with respect to \(u = -x^2\), and then multiplying by the derivative of \(u\), which is \(-2x\).
Mastering the Chain Rule is critical for success in calculus as it is frequently used in more complex differentiation problems.
Absolute Maximum and Minimum Values
To find the absolute maximum and minimum values of a function on a closed interval, it's vital to analyze the function at critical points and endpoints of that interval. These values show where a function realizes its largest and smallest outputs over a given domain, providing insight into the range and behavior of the function.
  • First, calculate the function's value at its critical points and at each endpoint of the interval. In our example, the function \(f(x) = e^{-x^2}\) is evaluated at \(x = 0\), \(x = -1\), and \(x = 1\).
  • After computation, you see that \(f(0) = 1\), \(f(-1) = e^{-1}\), and \(f(1) = e^{-1}\). Comparing these values allows you to conclude that the absolute maximum value is \(1\) at \(x = 0\), and the absolute minimum value is \(e^{-1}\) at the endpoints \(x = -1\) and \(x = 1\).
By evaluating these factors, you gain a clearer understanding of a function's behavior over an interval, critical for graphing and optimization problems in calculus.

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