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 4.1.37
Textbook Question
Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant.
ƒ(x) = x² √(x + 5)

1
To find the critical points of the function ƒ(x) = x² √(x + 5), we first need to find its derivative. Use the product rule for differentiation, since the function is a product of x² and √(x + 5).
The product rule states that if you have two functions u(x) and v(x), then the derivative of their product is u'(x)v(x) + u(x)v'(x). Here, let u(x) = x² and v(x) = √(x + 5).
Differentiate u(x) = x² to get u'(x) = 2x. Differentiate v(x) = √(x + 5) using the chain rule: v'(x) = (1/2)(x + 5)^(-1/2) * 1 = 1/(2√(x + 5)).
Apply the product rule: ƒ'(x) = 2x√(x + 5) + x²/(2√(x + 5)). Simplify the expression to make it easier to analyze.
Set ƒ'(x) = 0 to find the critical points. Solve the equation 2x√(x + 5) + x²/(2√(x + 5)) = 0 for x, considering the domain restrictions where x + 5 > 0.

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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 identifying local maxima, minima, and points of inflection. To find critical points, one typically differentiates the function and solves for the values of x that satisfy these conditions.
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Derivative
The derivative of a function measures the rate at which the function's value changes with respect to changes in its input. It is a fundamental concept in calculus, providing insights into the function's behavior, such as increasing or decreasing intervals. For the function given, applying the product and chain rules will be necessary to find its derivative.
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Derivatives
Product Rule
The product rule is a formula used to differentiate products of two functions. It states that if you have two functions u(x) and v(x), the derivative of their product is given by u'v + uv'. This rule is particularly relevant for the function in the question, as it involves the product of x² and √(x + 5).
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