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Chapter 3: Problem 56

A function \(f(x)\) is given. Find the \(x\) values where \(f^{\prime}(x)\) has a relative maximum or minimum. \(f(x)=e^{-x^{2}}\)

Short Answer

Expert verified
The derivative \(f'(x)\) has a relative maximum at \(x = 0\).

Step by step solution

01

Differentiate the Function

First, find the derivative of the function \(f(x) = e^{-x^2}\). Using the chain rule, we differentiate **\(e^{u}\)** where **\(u = -x^2\)**: \( f'(x) = e^{-x^2} \cdot (-2x) = -2xe^{-x^2} \).
02

Find Critical Points

Set the first derivative \(f'(x) = -2xe^{-x^2}\) equal to zero to find the critical points. Since \(e^{-x^2}\) is never zero, solve \(-2x = 0\), which gives **\(x = 0\)**.
03

Determine the Second Derivative

Find the second derivative, \(f''(x)\), to use the second derivative test. Differentiate \(f'(x) = -2xe^{-x^2}\) using the product rule: \[f''(x) = -2e^{-x^2} + 4x^2e^{-x^2} = 2x(2x - 1)e^{-x^2}\].
04

Evaluate the Second Derivative at Critical Points

Substitute the critical point \(x = 0\) into the second derivative to apply the second derivative test: \(f''(0) = 2 \cdot 0 \cdot (2 \cdot 0 - 1)e^{-0} = -2\).Since \(f''(0) < 0\), \(f(x)\) has a local maximum at \(x = 0\).

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Key Concepts

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

Chain Rule in Differentiation
The chain rule is a powerful technique in calculus used for finding the derivative of composite functions. A composite function means a function within another function, often noted as \( f(g(x)) \). In our original exercise, we had to differentiate \( f(x) = e^{-x^2} \), which is a good example of a composite function. Here, \( e^{u} \) is the outer function and \( u = -x^2 \) is the inner function.

To differentiate this, the chain rule tells us to:
  • Find the derivative of the outer function, leaving the inner function unaltered. For \( e^{u} \), the derivative is \( e^{u} \).
  • Then, multiply by the derivative of the inner function, \( u = -x^2 \) in this case. The derivative here is \(-2x\).
Therefore, applying the chain rule gives us \( f'(x) = e^{-x^2} \cdot (-2x) = -2xe^{-x^2} \). Understanding this rule is fundamental in handling more complex functions.
Second Derivative Test for Critical Points
The second derivative test is a method to determine whether a function has a local maximum or a local minimum at its critical points. A critical point occurs where the first derivative of the function is zero or undefined.

Once we find the critical points, we apply the second derivative test:
  • Calculate the second derivative of the function.
  • Evaluate it at the critical points found from the first derivative.
  • If the second derivative is positive (\( f''(x) > 0 \)), the function has a local minimum at that point.
  • If the second derivative is negative (\( f''(x) < 0 \)), the function has a local maximum.
In our exercise, after determining the critical point \( x = 0 \), the second derivative \( f''(0) \) was found to be \(-2\), indicating a local maximum at \( x = 0 \). This technique is essential for analyzing the behavior of functions near their critical points.
Understanding Differentiation
Differentiation is the process of finding the derivative of a function, which gives us the rate at which the function value is changing at any point. This is a central concept in calculus.

The derivative, noted as \( f'(x) \), can provide us:
  • The slope of the tangent to the graph of the function at any given point.
  • Information on increasing or decreasing behavior of functions.
  • Critical points where the derivative is zero or undefined, indicating potential local maxima or minima.
In our specific problem, differentiating \( f(x) = e^{-x^2} \) was key to identifying the critical point. Mastering differentiation helps us in dissecting the intricate behavior of function graphs.
Local Maximum in Function Analysis
A local maximum refers to a point on the graph of a function where the function value is higher than at neighboring points. This is distinct from a global maximum, where the function would be highest across its entire domain.

To determine a local maximum:
  • Identify critical points using the first derivative test, which involves setting the derivative equal to zero.
  • Use the second derivative test to classify these points further.
  • At a local maximum, the slope changes from positive to negative.
In our case, the function \( f(x) = e^{-x^2} \) has a local maximum at \( x = 0 \) because the second derivative was negative at this point \( f''(0) = -2 \). Recognizing local maxima is crucial in optimizing functions and understanding their peak performance points.

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Most popular questions from this chapter

A function \(f(x)\) is given. Find the \(x\) values where \(f^{\prime}(x)\) has a relative maximum or minimum. . \(f(x)=x^{2} \ln x\)

A function \(f(x)\) and interval \([a, b]\) are given. Check if Rolle's Theorem can be applied to fon \([a, b] ;\) if so, find cin \([a, b]\) such that \(f^{\prime}(c)=0\). \(f(x)=\cos x\) on \([0, \pi]\).

A function \(f(x)\) is given. (a) Give the domain of \(f\). (b) Find the critical numbers of \(f\). (c) Create a number line to determine the intervals on which \(f\) is increasing and decreasing. (d) Use the First Derivative Test to determine whether each critical point is a relative maximum, minimum, or neither. \(f(x)=\frac{x^{2}-4}{x^{2}-1}\)

A function \(f(x)\) and interval \([a, b]\) are given. Check if Rolle's Theorem can be applied to fon \([a, b] ;\) if so, find cin \([a, b]\) such that \(f^{\prime}(c)=0\). \(f(x)=x^{2}+x-6\) on [-3,2] .

A function \(f\) has derivative \(f^{\prime}(x)=(\sin x+2) e^{x^{2}+1},\) where \(f^{\prime}(x)>1\) for all \(x .\) Is \(f\) increasing, decreasing, or can we not tell from the given information?
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