Derivative word problem. Tricky

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SUMMARY

The discussion focuses on solving a derivative word problem involving a spaceship's trajectory defined by the cubic equation y = x³ - 8x. The goal is to determine the point P where the spaceship can reach the coordinates (4,0) after shutting off its engines. The slope of the tangent at point P is derived from the first derivative, m = 3x₁² - 8. The solution involves setting the equations of the cubic and the tangent line equal to each other to find the value of x₁.

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Homework Statement



A spaceship moves along y = x3 - 8x in the positive x-direction. Shutting off the engines at P allows it to move off along the tangent at P. Find P so that the ship can reach the point (4,0).

Homework Equations



Dunno

The Attempt at a Solution



So I start with P = (x1,y1). The slope of the tangent is then m = 3x12 - 8 from the first derivative.

Then I need P, so I went like this:

m = (y2 - y1) / (x2 - x1) = 3x12 - 8 where y2 and x2 are (4,0).

But now I have 1 Eqn. and 2 variables so I can't solve. I was thinking of using y = mx + b because I have y and x, plus an equation for m but then I end up with 2 equations and 3 variables because of the b term. Please help!
 
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Okay, m= 3x_1^2- 8 and you know, I presume, that a straight line (except vertical) can be written y= m(x-a)+ b where m is the slope and (a, b) is a point on the line. What is the equation of the line with slope 3x_1^2- 8 through (4, 0)?
You want that straight line to touch the graph when x= x1 so set the two equations, for the cubic and for the straight line equal and set x= x1. Solve for x1.
 
Ok! I got it! Thanks a bunch
 

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