What is the equation of the tangent at (3, f(3))?

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Discussion Overview

The discussion revolves around finding the equation of the tangent line to a curve at the point (3, f(3)). Participants explore the necessary steps and notation involved in this process, focusing on the application of derivatives and the equation of a line.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about the notation and the process of finding the tangent line, questioning the purpose of the coordinates.
  • Another participant clarifies that the task is to find the equation of the tangent at the specified point.
  • A participant suggests using the point-slope form of a line, indicating they have the necessary x and y values.
  • Another participant confirms the use of the formula y - f(c) = f'(c)(x - c) for the tangent line, specifying that c equals 3.
  • A later reply reiterates the formula and explains that the derivative at the point provides the slope of the tangent line.

Areas of Agreement / Disagreement

Participants generally agree on the formula to use for finding the tangent line, but there is some confusion regarding the notation and the steps involved in applying it.

Contextual Notes

There is a lack of clarity regarding the function f and its derivative, which may affect the application of the discussed formulas.

Joza
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I have a function.

I am asked to find the equation of the tangent to the curve at (3, f(3))

I think I am just confused of notation, but what is it asking here? To find the points, don't you put in the x value into the first derivative and this gets the y value, and use the formula for an equation of a line? What are those coordinates for then?
 
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Well, you're supposed to find the equation of the tangent at that very point.
 
So I already have my x and y, so just use y-y1=m(x-x1)?
 
If you meant y - f(c) = f'(c)(x - c), then yes. (c = 3)
 
radou said:
If you meant y - f(c) = f'(c)(x - c), then yes. (c = 3)

As you might know f'(c)=tg@, so the derivative of the curve at the point (c,f(c)) is actually the orientation coeficient of the tagent drawn at that point.

So this is actually what radou said, just giving u some more hints.
 

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