Gregory Ramos

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answers: 1

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Answer:

The function y = x^3 + x.

The equation for the tangents at the point where the slope is 4 is required.

The slope of the tangent at any point is equal to the value of the derivative y'.

y' = 3x^2+ 1

3x^2 + 1 = 4

=> 3x^2 = 3

=> x^2 = 1

=> x = 1 and x = -1

At x = 1, y = 2 and at x = -1, y = -2

The equation of the tangents are:

`(y - 2)/(x - 1) = 4`

=> y - 2 = 4x - 4

=> 4x - y - 2 = 0

And `(y + 2)/(x + 1) = 4`

=> y + 2 = 4x + 4

=> 4x - y + 2 = 0

The slope at any point is equal to 3x^2 + 1

This is least when 6x = 0

=> x = 0

**The value of x when the slope is the least is x = 0 and the least value of the slope is 1.**

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