Find the value of k for which y =2x + 1 is tangent to the curve y=x^3 +kx +3

and yes, the equation is y=x^3 +kx +3

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We want to find a k such that y=2x+1 is tangent to the graph of `y=x^3+kx+3`

First, since the line y=2x+1 is tangent to the graph, at the point of intersection of the line and the curve the slope of the curve, and its first derivative, have value 2.

`y=x^3+kx+3`

`y'=2=3x^2+kx ==> k=2-3x^2`

Also, at the point of tangency the line and curve intersect:

`2x+1=x^3+kx+3`

`x^3+(k-2)x+2=0` Substituting for k we get:

`x^3+[2-3x^2-2]x+2=0`

`x^3-3x^2+2=0`

`(x-1)(x^2-2x-2)=0`

So we have three points of intersection -- at only one will the line be tangent to the curve.

If x=1 then k=-1; the slope of the curve is 2. Thus the line is tangent to the curve at (1,3).

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When k=-1 then the line y=2x+1 will be tangent to `y=x^3+kx+3`

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The graph of `y=2x+1,y=x^3-x+3`

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