Coordinates at which 2 tangent lines of a function pass through a single point

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Discussion Overview

The discussion revolves around finding the coordinates of points on the graph of the function y=3e^x - x^3 where the tangent lines at those points pass through the point (1,0). The scope includes mathematical reasoning and problem-solving related to derivatives and tangent lines.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant outlines the problem and suggests using the coordinates of the tangent points and the point (1,0) to calculate the slope of the tangent line.
  • Another participant questions which point the derivative should be evaluated at to find the slope of the tangent line.
  • A third participant asks for the equation of the tangent line at a general point (a, f(a)), indicating the complexity of the final expression to solve.
  • A participant shares the derived equation 6e^x + 2x^3 - 3x^2 - 3xe^x = 0 and expresses frustration about solving it, while also mentioning that the book's answers satisfy this equation.
  • One participant suggests using a graphing calculator to find x-intercepts of the derived equation.
  • A later reply mentions Newton's Method as a potential approach for solving the equation.

Areas of Agreement / Disagreement

The discussion does not reach a consensus on the best method to solve the equation or the steps involved in finding the tangent points, indicating that multiple approaches and uncertainties remain.

Contextual Notes

The final expression derived is noted to be complex, and there are indications that numerical methods or computational tools may be necessary for evaluation. The discussion also reflects varying levels of comfort with mathematical techniques among participants.

goaliemon89
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Here is what the question asks:

Find the coordinate of all points in the graph of y=3e^x - x^3 at which the tangent line passes through the point (1,0).

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I am told that the 2 points are (-.872, -1.027) and (2.275, 13.657). Any help at all that you can give will be very much appreciated.

I was thinking of letting the 2 coordinates be (X1, Y1) and (X2, Y2), calculating the slope of the line through (1,0) and one of these points, and setting this slope equal to the derivative of y=3e^x - x^3 ...but I seem to be going no where.

Suggestions?
 
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Yes, the slope of the derivative would be the slope of the tangent at what point on the line?
 
What's the equation for the tangent line at the point (a,f(a))?

By the way. This final expression is nasty to solve. I think you'll need a computer to evaluate the answers numerically in the end.
 
Alright -- the final equation i get is 6e^x + 2x^3 - 3x^2 - 3xe^x = 0

The 2 answers that the book gives work in this equation...but how the hell would I solve this?

If I were to set 6e^x + 2x^3 - 3x^2 - 3xe^x equal to y in my calculator and graph it, is there a way for the calculator to tell you the x-intercepts?

EDIT - Never mind - I figure it out! Thanks for helping out.
 
Last edited:
Newton's Method?
 

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