Question at position 9 Test the following for relative maxima and minima for: y=18x−23x3y=18x-\frac{2}{3}x^3.Relative maximum at x = 0, relative minimum at x = +3Relative maximum at x = +3, relative minimum at x = −3No relative maximum, relative minimum at x = -3Relative maximum at x = -3, relative minimum at x = +3

Question

Question at position 9 Test the following for relative maxima and minima for: y=18x−23x3y=18x-\frac{2}{3}x^3.Relative maximum at x = 0, relative minimum at x = +3Relative maximum at x = +3, relative minimum at x = −3No relative maximum, relative minimum at x = -3Relative maximum at x = -3, relative minimum at x = +3

BlackTom AI · Accepted Answer

First, rewrite the problem in my own words to ensure clarity: we have y = 18x − (2/3)x^3, and we need to locate relative extrema by finding critical points and applying the second derivative test. Option A: Relative maximum at x = 0, relative minimum at x = +3. This would require that x = 0 be a critical point and yield a local max, and x = +3 yield a local min. However, the first derivative y' = 18 − 2x^2 is zero when x^2 = 9, i.e., x = ±3. Therefore, x = 0 is not a critical point at all, so this option cannot be correct. Option B: Relative maximum at x = +3, relative minimum at x = −3. This matches the setup where the critical points are at x = ±3. To decide which is max or min, apply the second derivative test: y'' = −4x. At x = +3, y'' = −12 < 0, indicating a local maximum there. At x = −3, y'' = +12 > 0, indicating a local minimum there. Hence this option is consistent with the second-derivative results. Option C: No relative maximum, relative minimum at x = −3. This claims there is no maximum at any point, which would require y'' ≥ 0 for all critical points or some other reasoning, but since we do have a critical point at x = +3 with y'' < 0, there is a local maximum there. Therefore this option misstates the situation. Option D: Relative maximum at x = −3, relative minimum at x = +3. This reverses the roles determined by the second derivative test: at x = −3, y'' = +12 > 0 would indicate a minimum, not a maximum; at x = +3, y'' = −12 < 0 indicates a maximum, so the labeling in this option is incorrect. To summarize the reasoning steps: compute y' to locate critical points, solve 18 − 2x^2 = 0 to find x = ±3; compute y'' = −4x to classify each critical point; since y''(−3) > 0, x = −3 is a relative minimum, and since y''(3) < 0, x = 3 is a relative maximum. The option that aligns with these conclusions is the one stating a relative maximum at x = +3 and a relative minimum at x = −3.

View Explanation

Log in for full answers

Similar Questions

Question text 1. Answer 1 Question 10[input] 2. Answer 2 Question 10[input]

MTH1010_10_01_2

Let be the function defined on whose graph is given below, and let . Which of the following statements is true?

Let and . What can you say about ?

Let be a function such that is continuous on , , and . Consider . Which of the following can be determined about at ?

Let be a function such that is continuous on , , and . Consider . Which of the following can be determined about at ?

Find the value of x at which the following function has a local minimum: 𝑓 ( 𝑥 ) = 𝑥 ( 𝑥 2 − 5 ) 1 3

Find the value of x at which the following function has a local minimum: f\left(x\right)=\frac{x}{\left(x^2-5\right)^{\frac{1}{3}}}

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler