Finding the Slope of a Tangent Line to f(x) at (2,3)

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Homework Help Overview

The problem involves finding the slope of the tangent line to the function f(x) = x - 1/x at the point (2,3). The discussion centers around the concept of derivatives and their application in determining the slope of tangent lines.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definition of the derivative as the slope of the tangent line and explore how to differentiate the given function. There are questions about the relevance of the specific point (2,3) to the slope calculation and the implications of the function's value at that point.

Discussion Status

Some participants have provided insights into the derivative's role in finding the slope, while others have noted potential discrepancies regarding the point of tangency. The conversation reflects an exploration of the definitions and implications of the tangent line's slope without reaching a consensus.

Contextual Notes

There is a mention of a possible typo regarding the point (2,3) not being on the graph of the function, which raises questions about the assumptions made in the problem setup.

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Homework Statement



the line tangent to the graph of f(x) = x-1/x at (2,3) has what slope?

The Attempt at a Solution



Same story with this problem, I remember doing it but can't remember how to attack the problem. can someone give me a good first step to go off of?
 
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The slope of the tangent line to a curve, at any point, is the value of the derivative their. That's often taken as a definition of derivative. In order to differentiate, I would write that as f(x)= x- x-1.
 
1 + 1/4 = 5/4, gotcha.

so the 3/2 on the y-axis has no role in determining the final answer?
 
Of minor note, the graph doesn't actually pass through (2,3), but I'll assume it's a typo (though it would be a mean trick question)

The value of f(x) is pretty much irrelevant to the slope of the tangent line, yes. (in fact, note you can add any constant to f, making it f(x)+c, and it will still have the same derivative.)
 

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