Homework Help is Here – Start Your Trial Now!
Math
Calculus
Show analytically (that is, without reference to a graph) that h(x) = v1 + x is always increasing, given that Dom h = [1,

Show analytically (that is, without reference to a graph) that h(x) = v1 + x is always increasing, given that Dom h = [1,

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
See similar textbooks
icon
Related questions
Question
**Problem Statement:** Show analytically (that is, without reference to a graph) that \( h(x) = \sqrt{1 + x^3} \) is always increasing, given that \( \text{Dom } h = [1, \infty) \). --- To demonstrate that the function \( h(x) \) is always increasing, you'll need to analyze its derivative \( h'(x) \). A function is increasing on an interval if its derivative is positive for all values in that interval. 1. **Differentiate \( h(x) \):** Apply the chain rule: \[ h(x) = (1 + x^3)^{1/2} \] \[ h'(x) = \frac{1}{2}(1 + x^3)^{-1/2} \cdot 3x^2 = \frac{3x^2}{2\sqrt{1 + x^3}} \] 2. **Analyze the derivative \( h'(x) \):** For the domain \( x \in [1, \infty) \): - The numerator, \( 3x^2 \), is always positive since \( x^2 \ge 0 \) and multiplying by 3 keeps it positive over the domain. - The denominator, \( 2\sqrt{1 + x^3} \), is always positive as well, since \( \sqrt{1 + x^3} \) is the square root of a positive function. Thus, \( h'(x) = \frac{3x^2}{2\sqrt{1 + x^3}} > 0 \) for all \( x \ge 1 \). 3. **Conclusion:** Since \( h'(x) > 0 \) for all \( x \) in the domain \([1, \infty)\), the function \( h(x) = \sqrt{1 + x^3} \) is always increasing on this interval.
Transcribed Image Text:**Problem Statement:** Show analytically (that is, without reference to a graph) that \( h(x) = \sqrt{1 + x^3} \) is always increasing, given that \( \text{Dom } h = [1, \infty) \). --- To demonstrate that the function \( h(x) \) is always increasing, you'll need to analyze its derivative \( h'(x) \). A function is increasing on an interval if its derivative is positive for all values in that interval. 1. **Differentiate \( h(x) \):** Apply the chain rule: \[ h(x) = (1 + x^3)^{1/2} \] \[ h'(x) = \frac{1}{2}(1 + x^3)^{-1/2} \cdot 3x^2 = \frac{3x^2}{2\sqrt{1 + x^3}} \] 2. **Analyze the derivative \( h'(x) \):** For the domain \( x \in [1, \infty) \): - The numerator, \( 3x^2 \), is always positive since \( x^2 \ge 0 \) and multiplying by 3 keeps it positive over the domain. - The denominator, \( 2\sqrt{1 + x^3} \), is always positive as well, since \( \sqrt{1 + x^3} \) is the square root of a positive function. Thus, \( h'(x) = \frac{3x^2}{2\sqrt{1 + x^3}} > 0 \) for all \( x \ge 1 \). 3. **Conclusion:** Since \( h'(x) > 0 \) for all \( x \) in the domain \([1, \infty)\), the function \( h(x) = \sqrt{1 + x^3} \) is always increasing on this interval.
Expert Solution
Step 1

Calculus homework question answer, step 1, image 1

steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Knowledge Booster
Asymptote
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS