Question on the slope of the tangent

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Homework Help Overview

The original poster is investigating the points on the graph of the function y = (1/3)x^3 - 5x - (4/x) where the slope of the tangent is horizontal. This involves concepts from calculus, specifically derivatives and their relationship to the slope of a curve.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to find the derivative of the function and set it equal to zero to determine where the slope of the tangent is horizontal. There are questions about the necessity of using specific formulas, such as the quotient difference formula, and how to relate the derivative to the slope of a horizontal line.

Discussion Status

The discussion includes various suggestions and hints regarding the approach to finding the derivative and understanding the concept of horizontal tangents. Some participants emphasize the importance of setting the derivative to zero, while others clarify the relationship between the derivative and the slope of a horizontal line.

Contextual Notes

There is a mention of potential constraints regarding the use of specific formulas in the problem-solving process, as well as a focus on understanding the fundamental concepts of derivatives and slopes.

kiss89
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Find the points on the graph of y= (1/3)x^3-5x- (4/x) at which the slope of the tangent is horizontal.

what i know:
- we have to use m=[f(a+h)-f(a)]/h
- if we change the equation we can get 3x^4 - 15x^2 -12
- the slope of the tangent is zero.

THANX
 
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hint: how does the derivative of a function relate to the slope of the tangent of its graph?
 
Find the derivative, and set it equal to the slope of a horizontal line. What is that?
 
What everyone is saying is "DO IT"! By the way are you really required to use the "quotient difference" formula? It's not too difficult but tedious and most problems like this allow the use of derivative formulas.
 
Gib Z said:
Find the derivative, and set it equal to the slope of a horizontal line. What is that?

Do you mean, "what is the slope of a horizontal line?"

Draw a graph that shows a horizontal line. Pick two points [itex](x_1,y_1)[/itex] and [itex](x_2,y_2)[/itex] on the line. Do you know how to calculate the slope of a line from two points?
 
yes that's what i meant, but i knew the answer..set the derivative to zero is what i meant
 

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