Math

Find 2 tangent line equations to the curve y=x^3+x at the points where the slope of the curve is 4.

What is the smallest possible slope of the curve?

At what x-value(s)does the curve have this slope?


So far I have figured out the derivative of this function is
y'= 3x^2+1.

I think one of the equations is y-2 = 4(x-1).
I think smallest slope is 4.
I think x=1 and x=-1.

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asked by Dave
  1. for the two tangent lines, where is 3x^2+1 = 4?
    x = ±1
    y(1)=2
    y(-1) = -2
    So, the two lines are
    y-2 = 4(x-1)
    y+2 = 4(x+1)

    y" = 6x
    so, at x=0 the slope is smallest: 1

    If you look at the graph, you can clearly see that the slope at x=0 is smaller than at x=1 or -1.

    http://www.wolframalpha.com/input/?i=x%5E3%2Bx

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    posted by Steve

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