Domanda del libro di testo
17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.
lim_x→ 0 (3 sin 4x) / 5x
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Ch. 4 - Applications of the Derivative
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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TOC.tableOfContents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 4, Problema 4.1.39
Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant.
ƒ(x) = x √(x-a)
Guida verificata passo dopo passo1
To find the critical points of the function \( f(x) = x \sqrt{x-a} \), we first need to find its derivative. Start by expressing the function in a form that is easier to differentiate: \( f(x) = x (x-a)^{1/2} \).
Apply the product rule to differentiate \( f(x) = x (x-a)^{1/2} \). The product rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then \( (uv)' = u'v + uv' \). Here, let \( u(x) = x \) and \( v(x) = (x-a)^{1/2} \).
Differentiate \( u(x) = x \) to get \( u'(x) = 1 \). Differentiate \( v(x) = (x-a)^{1/2} \) using the chain rule: \( v'(x) = \frac{1}{2}(x-a)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x-a}} \).
Substitute the derivatives back into the product rule: \( f'(x) = 1 \cdot (x-a)^{1/2} + x \cdot \frac{1}{2\sqrt{x-a}} \). Simplify this expression to find \( f'(x) \).
Set \( f'(x) = 0 \) to find the critical points. Solve the equation for \( x \) to determine the values where the derivative is zero or undefined, which will give the critical points of the function.
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Critical Points
Critical points of a function occur where its derivative is either zero or undefined. These points are essential for determining local maxima, minima, and points of inflection. To find critical points, one must first compute the derivative of the function and then solve for the values of x that satisfy the condition of the derivative being zero or undefined.
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Critical Points
Derivative
The derivative of a function measures the rate at which the function's value changes as its input changes. It is a fundamental concept in calculus that provides information about the function's slope at any given point. For the function ƒ(x) = x √(x-a), applying the product and chain rules will be necessary to find its derivative.
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Percorso guidato
05:44Derivatives
Function Behavior
Understanding the behavior of a function involves analyzing its continuity, limits, and the nature of its critical points. This includes determining whether critical points correspond to local maxima, minima, or saddle points. By evaluating the second derivative or using the first derivative test, one can gain insights into the function's overall shape and trends.
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5:46Graphs of Exponential Functions
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