Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 2h 22m
5. Graphical Applications of Derivatives
Intro to Extrema
Problem 4.1.39
Textbook Question
Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant.
ƒ(x) = x √(x-a)

1
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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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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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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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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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Graphs of Exponential Functions
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