Two different tangents which are perpendicular?

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SUMMARY

The discussion focuses on determining whether the curves y=x^2 and y=x^3 can have two different tangents that are perpendicular to each other. For the quadratic curve y=x^2, the derivative is (x^2)'=2x, leading to the condition that the slopes of two tangents, m1 and m2, must satisfy m1 * m2 = -1. For the cubic curve y=x^3, the derivative is (x^3)'=3x^2, resulting in the equation (3x1^2)(3x2^2) = -1. The conclusion is that both curves can have perpendicular tangents under specific conditions.

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Homework Statement


Does the curve y=x^2 have two different tangents which are perpendicular? Does the curve y=x^3?

The Attempt at a Solution


I have no idea how to prove that for x^2 or x^3.
(x^2)' =2x that is tangent at a point x
then m = -1/2 x for another point...
How do i prove that there is sucha point? then how about x^3?
 
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Two lines are perpendicular if and only if the product of their slopes is -1. So the question is, "are the two different values of x, say x1 and x2, such that (2x1)(2x2)= -1?"
For the cubic, the corresponding equation is (3x12)(3x22)= -1.
 

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