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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 4 - Applications of the DerivativeProblem 4.4.46
Chapter 4, Problem 4.4.46

Graphing functions Use the guidelines of this section to make a complete graph of f.


f(x) = e⁻ˣ²/₂

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Identify the function: The given function is \( f(x) = e^{-x^2/2} \). This is a Gaussian function, which is a type of bell curve centered at the origin.
Determine the domain and range: The domain of \( f(x) \) is all real numbers, \( (-\infty, \infty) \), because the exponential function is defined for all real numbers. The range is \( (0, 1] \) because \( e^{-x^2/2} \) is always positive and reaches a maximum value of 1 when \( x = 0 \).
Find the symmetry: The function \( f(x) = e^{-x^2/2} \) is an even function because \( f(-x) = f(x) \). This means the graph is symmetric with respect to the y-axis.
Determine the critical points: To find critical points, compute the derivative \( f'(x) \) and set it to zero. The derivative is \( f'(x) = -x e^{-x^2/2} \). Setting \( f'(x) = 0 \) gives \( x = 0 \) as the only critical point. Since \( f'(x) \) changes sign around \( x = 0 \), this is a maximum point.
Analyze the behavior at infinity: As \( x \to \pm\infty \), \( e^{-x^2/2} \to 0 \). This means the graph approaches the x-axis but never touches it, indicating horizontal asymptotes at \( y = 0 \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Exponential Functions

Exponential functions are mathematical functions of the form f(x) = a^x, where 'a' is a constant. In the given function f(x) = e^(-x²/2), 'e' is the base of the natural logarithm, approximately equal to 2.71828. Understanding the behavior of exponential functions, especially with negative exponents, is crucial for graphing as they determine the function's growth or decay.
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Transformation of Functions

Transformations involve shifting, stretching, or compressing the graph of a function. For f(x) = e^(-x²/2), the exponent -x²/2 indicates a horizontal reflection and a vertical compression. Recognizing these transformations helps in predicting the shape and orientation of the graph, which is essential for accurate plotting.
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Critical Points and Symmetry

Critical points, where the derivative is zero or undefined, help identify local maxima, minima, or points of inflection. For f(x) = e^(-x²/2), symmetry about the y-axis is evident due to the even power of x. Analyzing these aspects allows for a more complete understanding of the function's behavior and assists in creating a detailed graph.
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Related Practice
Textbook Question

{Use of Tech} Special curves The following classical curves have been studied by generations of mathematicians. Use analytical methods (including implicit differentiation) and a graphing utility to graph the curves. Include as much detail as possible.


x⁴ - x² + y² = 0 (Figure-8 curve)

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Textbook Question

Finding antiderivatives. Find all the antiderivatives of the following functions. Check your work by taking derivatives.


g(s) = 1 / (s² + 1)

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Textbook Question

Suppose f is differentiable on (-∞,∞), f(1) = 2, and f'(1) = 3. Find the linear approximation to f at x = 1 and use it to approximate f (1.1).

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Textbook Question

Maximum-area rectangles Of all rectangles with a perimeter of 10, which one has the maximum area? (Give the dimensions.)

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Textbook Question

Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.


ƒ(x) = x/(x²+9)⁵ on [-2,2]

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Textbook Question

{Use of Tech} Finding roots with Newton’s method For the given function f and initial approximation x₀, use Newton’s method to approximate a root of f. Stop calculating approximations when two successive approximations agree to five digits to the right of the decimal point after rounding. Show your work by making a table similar to that in Example 1.


f(x) = x² - 10; x₀ = 3

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