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:

$$y = 6x +/- 2$$

  • $\begingroup$ As @amd said, you're not stuck -- you're done! :-) $\endgroup$ – zipirovich Apr 5 '18 at 18:51

You’ve already solved the problem: there are two such tangent lines. The function is well-behaved and has a tangent at each point, and you’ve determined that there are two points at which the derivative has the right value. There’s no need to go any further with this unless there’s a follow-up question that asks you to produce the equations of those lines.

  • $\begingroup$ Yes, that should have made sense to me... there are two solutions to f'(x) = 6 so there are 2 lines tangent to f(x) and parallel to y=6x+1. There is indeed a follow-up question that asks to produce the equations of those lines. I just didn't want to over rely on MSE because I'm well aware MSE has a problem with people coming here asking other people to do their homework. $\endgroup$ – Victor S. Apr 5 '18 at 19:35

Note: The question asks only for how many so the answer is immeediately two. For the actual equations of the tangents:

The derivative is $$3x^2+3=6\implies x=\pm1$$ The points are therefore $(-1,-4),(1,4)$ - substitute into $y=x^3+2x$.

Let's consider the first point. For a tangent line parallel to $y=6x+1$, the slope must be $6$. So we have that $$y=mx+c\implies-4=6(-1)+c\implies c=2\implies \boxed{y=6x+2}$$

Can you do the other one?

  • $\begingroup$ f(x) = x³ + 3x, not x³ - 3x, sorry. $\endgroup$ – Victor S. Apr 5 '18 at 15:53
  • $\begingroup$ No problem, edited! $\endgroup$ – TheSimpliFire Apr 5 '18 at 18:42
  • $\begingroup$ The equation of the tangents is the follow-up question but I didn't want to have MSE do my homework as i'm well aware this is a big problem here... Check my edit on the OP. $\endgroup$ – Victor S. Apr 5 '18 at 19:37

enter image description here

Plot of the original function and the equation of the lines tangent to f with slope = 6


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