Skip to main content
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 4.1.37
Chapter 4, Problem 4.1.37

Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant.


ƒ(x) = x² √(x + 5)

Verified step by step guidance
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.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6m
Was this helpful?

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.
Recommended video:
04:50
Critical Points

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.
Recommended video:
05:44
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).
Recommended video:
05:18
The Product Rule
Related Practice
Textbook Question

Explain the Mean Value Theorem with a sketch.

Textbook Question

{Use of Tech} Tumor size In a study conducted at Dartmouth College, mice with a particular type of cancerous tumor were treated with the chemotherapy drug Cisplatin. If the volume of one of these tumors at the time of treatment is V₀, then the volume of the tumor t days after treatment is modeled by the function V(t) = V₀ (0.99e⁻⁰·¹²¹⁶ᵗ + 0.01e⁰·²³⁹ᵗ). (Source: Undergraduate Mathematics for the Life Sciences, MAA Notes No. 81, 2013)


Plot a graph of y = 0.99e⁻⁰·¹²¹⁶ᵗ + 0.01e⁰·²³⁹ᵗ, for 0 ≤ t ≤ 16, and describe the tumor size over time. Use Newton’s method to determine when the tumor decreases to half of its original size.

Textbook Question

Particular antiderivatives For the following functions f, find the antiderivative F that satisfies the given condition.


f(v) = sec v tan v; F(0) = 2, -π/2 < v < π/2

Textbook Question

Minimum distance Find the point P on the line y = 3x that is closest to the point (50, 0). What is the least distance between P and (50, 0)?

Textbook Question

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.


lim_x→0⁺ x²ˣ

Textbook Question

Give the antiderivatives of xᵖ . For what values of p does your answer apply?

1
views