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  • Question

    Find the point where the curves $$\tag 1y = x^3 -3x + 4$$ and $$\tag 2 y = 3x^2 - 3x$$ are tangent to each other, that is, have a common tangent line.


  • My approach

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    • Let $x = a$ and $x = b$ be the points on curves $(1)$ and $(2)$, respectively, at which their slopes are equal and share a mutual tangent line.

    • Now I will relate the $a$ and $b$ by equating the derivatives of $(1)$ at $a$ and $(2)$ at $b$, as follows $$3a^2 - 3 = 6b- 3 \Leftrightarrow b = \frac{a^2}{2}$$

    • Let $A$ be the point on curve $(1)$ and $B$ be the point on curve $(2)$ where the two curves share the mutual tangent, that is $$A(a, x^3 - 3x + 4)$$ and $$B(b, 3b^2 - 3b) = B\Big(\frac{a^2}{2}, \frac{3a^4 - 6a^2}{4}\Big)$$

    • Now, since I have two points on the tangent line, I can calculate the slope and equate it to the derivative of $(1)$ at $a$ as follows $$\frac{(x^3 - 3x + 4) - \Big(\frac{3a^4 - 6a^2}{4}\Big)}{a - \frac{a^2}{2}} = 3a^2 - 3$$

    • Simplifying that equation I get the following, $$3a^4 -8a^3 - 12a^2 + 16 = 0$$

Now, I would solve for $a$ and then substitute the value of $a$ in to points $A$ and $B$ which would then be the points at which the two curves have a common tangent line. The problem is that I doubt I should be solving such an equation, and quite frankly, I don't have the tools to solve that equation, unless I'm missing something?

Any suggestions?

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  • $\begingroup$ Does "have a common tangent line" require that the two curves intersect each other as well? We can imagine a line that is tangent to both curves without the points of tangency being identical. $\endgroup$ – Rory Daulton Nov 10 '14 at 22:38
  • $\begingroup$ @RoryDaulton that's what caused me some confusion, glad you ask. Prior to doing this question I did a question where two parabolas (parabolas that don't intersect at all) share two tangent lines for which I had to find the coordinates at which these tangents touch the two parabolas. However, as Adriano, who provided a solution, pointed out that "have a common tangent line" means that they intersect and at the point of intersection they have the same slope. So I don't know why this book would be so, seemingly, ambiguous. What do you think? $\endgroup$ – Kermit the Hermit Nov 10 '14 at 22:46
  • $\begingroup$ I found this problem in James Stewart Calculus, Problems Plus $\endgroup$ – Joao Noch Oct 28 '17 at 7:35
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Two curves $y = f(x)$ and $y = g(x)$ have a common tangent line at $x = a$ iff:

  • They intersect there: $f(a) = g(a)$.
  • Their tangent lines have equal slope there: $f'(a) = g'(a)$.

Since quadratic equations are easier to solve than cubic ones, we start with the second condition: $$ 3a^2 - 3 = 6a - 3 \iff a^2 - 2a = 0 \iff a = 0, 2 $$ We now check if each candidate satisfies the first condition: $$ f(0) = 4 \neq 0 = g(0) $$ but: $$ f(2) = 6 = g(2) $$ So the only common tangent line occurs at $x = 2$ and is given by: $$ y - 6 = 9(x - 2) $$

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  • $\begingroup$ I entirely misunderstood the question. Prior to doing this question I did a question where two parabolas (parabolas that don't intersect at all) share two tangent lines for which I had to find the coordinates at which these tangents touch the two parabolas. $\endgroup$ – Kermit the Hermit Nov 10 '14 at 22:40
  • $\begingroup$ Can you find such tangent in this question? Thank you for taking the time clear up the confusion, I appreciate it. $\endgroup$ – Kermit the Hermit Nov 10 '14 at 22:42
  • $\begingroup$ Ah I see the confusing part. If $a \neq 2$ so that you are not dividing by zero in your slope formula expression, then you should instead obtain: $$ 0 = 3 a^4-8 a^3+16 = (a-2)^2 (3 a^2+4 a+4) $$ Since $(3 a^2+4 a+4) \geq 4 > 0$, we conclude that $a = 2$ is the only real solution. $\endgroup$ – Adriano Nov 10 '14 at 22:56
  • $\begingroup$ Oh man, I made a mistake with the simplifying of the slope formula. How did you factor $3a^4 - 8a^3 + 16$? Why do you say $3a^4 - 8a^3 + 16 \geq 4$? From graphing $3a^4 - 8a^3 + 16$, it seems greater or equal to approximately $2.6$? $\endgroup$ – Kermit the Hermit Nov 10 '14 at 23:48
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    $\begingroup$ I see, I feel comfortable with the question now. You've been a scholar and a gentleman. $\endgroup$ – Kermit the Hermit Nov 11 '14 at 0:11

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