I'm getting stuck with this problem:
How many lines tangent to the graph of $f(x)=x^3+3x$ are parallel to the line $y=6x+1$?
What I've done is this:
Since we're looking for lines parallel to $y=6x+1$ then obviously we're looking for lines of the form $y=6x+b$, so the real question is finding the values of $b$ that satisfy the conditions.
Since the lines are tangent to $f$ we need the derivative of $f$, that is $f'(x)=3x^2+3$.
Since the derivative of $f$ represents the slope of line tangent to $f$, we need $f'(x)=3x^2+3=6$ and we get the solutions $x=1$ and $x=-1$.
And this is where I'm stuck. There's some insight that I'm not having, I guess.
There is a follow-up question asking me to produce the equations of those two lines. I didn't want to share because I'm aware of people abusing MSE for their homework. This question was created by the Math department of my university and the official answer is $$y = 6x + 2$$ and $$y = 6x - 2$$
I asked my tutor, and he arrived at the official answers this way:
$$x³ + 3x = 6x + b => x³ - 3x = b$$
Then you plug in the $+/- 1$ found earlier to get $b = +/-2$ and finally arrive at the official answer for the equations of the lines:
$$y = 6x +/- 2$$