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Using the first and second derivative conditions, find the relative maximum(s) and minimum(s) of the function y = x³ – 15x² – 33x.

Using the first and second derivative conditions, find the relative maximum(s) and minimum(s) of the function y = x³ – 15x² – 33x.

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...
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**Problem Statement:** Using the first and second derivative conditions, find the relative maximum(s) and minimum(s) of the function \[ y = x^3 - 15x^2 - 33x. \]
Transcribed Image Text:**Problem Statement:** Using the first and second derivative conditions, find the relative maximum(s) and minimum(s) of the function \[ y = x^3 - 15x^2 - 33x. \]
The average cost \(\bar{c}\) for producing \(q\) units of a product is given by the formula: \[ \bar{c} = 0.01q^2 + 11 + \frac{1000}{q} \] To find the marginal cost when \(q = 10\), we need to differentiate this formula with respect to \(q\) and then substitute \(q = 10\) into the derivative. The marginal cost is the derivative of the average cost function, representing the rate of change of the total cost with respect to the quantity produced.
Transcribed Image Text:The average cost \(\bar{c}\) for producing \(q\) units of a product is given by the formula: \[ \bar{c} = 0.01q^2 + 11 + \frac{1000}{q} \] To find the marginal cost when \(q = 10\), we need to differentiate this formula with respect to \(q\) and then substitute \(q = 10\) into the derivative. The marginal cost is the derivative of the average cost function, representing the rate of change of the total cost with respect to the quantity produced.
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