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

Question

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.

Answer 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