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I have the following question:

Prove that the line $y_1=9x+17$ is tangent to the graph of the function $y_2=x^3-3x+1$. Find the point of tangency.

So, what I did was:


Let's construct a function $h$, such that $$h(x)=x^3-3x+1-(9x+17)=x^3-12x-16.$$

This function is continuous everywhere, particularly in the interval $[0,5]$. Now, $$h(0)=(0)^3-12(0)-16=-16$$ and $$h(5)=(5)^3-12(5)-16=49.$$ So, by Bolzano's theorem, there must exist a point $c \in (0,5)$ such that $h(c)=0$. This proves that the line $y_1$ and the function $y_2$ are tangent to each other at least at one point.

Now, we can think of the point of intersection as the point in which $h'(x)=0$. So, $$h'(x)=0 \iff 3x^2-12=0 \iff x \in \{-2,2\}.$$ Now, if either one of those two points is the point of intersection, then it must be a root of $h(x)$, so plugging them in we have $$h(2)= (2)^3-12(2)-16 = -32 \neq 0$$ and $$h(-2)=(-2)^3-12(-2)-16=24-24=0.$$ Therefore, $x=-2$ is the point of intersection of $y_1$ with $y_2$.


Now, my issue here is that I don't know exactly how to justify that what I did was actually right. In particular, I don't know why taking the difference of both original functions (constructing $h$) is a valid move and why is it that finding the points in which $h'(x)=0$ equates to finding the point of intersection between $y_1$ and $y_2$. So, my question is: if what I did was actually correct, why is it so? And if it's not, why am I wrong?.

Thanks in advance.

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  • $\begingroup$ @MatthewLeingang Thanks for your feedback, Matthew. I know that using the IVT is maybe a little bit of an overkill for this particular problem, but still, I want to be able to explain why using it leads to a correct answer. I've checked things graphically and my answer seems be correct, but if this question came up in a test, I wouldn't be able to justify my answer properly. That's why I'm asking for help. $\endgroup$
    – Eduardo M.
    Apr 20 '16 at 23:54
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I have a comment, and I know this has been looked at:

The statement involving Bolzano's Theorem, where you find that there is $c\in (0,5)$ where $h(c)=0$, implies that the graphs of the two functions intersect. It doesn't imply tangency, and it's actually a distraction, in that it gives you a point in $(0,5)$, which is quite far from the point of tangency you eventually find. (In fact, if you try using Bolzano's Theorem near the point of tangency, you must look at $h'(x)$, and not $h(x)$, since $x=-2$ gives rise to a local maximum of $h(x)$, where the function $h(x)$ is tangent to the $x$-axis.)

If you want to use Bolzano's Theorem, here's one (overly complicated) approach. First, notice that

$$h'(-3)=15,\ \ h'(0)=-12,\ \ h'(3)=15$$

Then by Bolzano's Theorem, there exists $c\in (-3,0)$ so that $h'(c) = 0$, and there also exists $c'\in (0,3)$ so that $h'(c) = 0$. Since $h'(x)$ is a quadratic, you can find the two solutions to the equation $h'(x)=0$ by hand, namely at $c'=2$, $c=-2$ (with notation consistent with our conclusion from Bolzano's Theorem). Since it's a quadratic, we know that we have accounted for all of them. (As the other answer states, this application of Bolzano's Theorem actually serves as an unnecessary extra step, since we can find the zeroes of $h'(x)$ by hand.)

You still need the two curves to intersect at the point to have a point of tangency, which is when we use the zeroes of $h'(x)$ (the critical numbers) and check which of these are zeroes of $h(x)$. In checking, you find that $h(2) = -32$, while $h(-2) = 0$. Now plug $x=-2$ back into your starting functions $y_1$ (or $y_2$) in order to conclude that you have a single point of tangency at $(-2,-1)$.


Of course, the less complicated alternative is to check that $h(x)$ has a double-root at $x=-2$, which it does, since $h(x) = (x+2)^2(x-4)$, which guarantees a point of tangency there. (This method is exactly that of the other answer, where you find roots of $h(x)$, and then check to see which of those is a root of $h'(x)$.)

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  • $\begingroup$ I see now that using Bolzano's Theorem first to check for intersection is redundant, I had failed to notice that finding the actual point of tangency in $h(x)$ also implied intersection. I guess I was doing things mechanically. Thanks! $\endgroup$
    – Eduardo M.
    Apr 21 '16 at 1:44
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    $\begingroup$ There isn't anything wrong with doing things mechanically, but you do have to exercise caution when doing this. Sometimes, if a step in your work gives a conclusion that either doesn't seem relevant or doesn't serve to answer the question, go back one step and see if maybe there isn't information you're missing, or if there might be a different approach that accomplishes what you want. $\endgroup$ Apr 21 '16 at 1:53
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    $\begingroup$ Basically, when doing things mechanically, make sure you are always keeping the end goal in mind. At each step, ask yourself: does this get me closer to the end goal? If not, why not? If so, how do I proceed forward from here? And if the answer is "I don't know", think about it for a while and see if you can't break it down further. It won't always lead to the most elegant or elementary proofs, but keeping a general outline or road map in mind can make proofs easier to write. $\endgroup$ Apr 21 '16 at 1:56
  • $\begingroup$ As a beginner, this is very valuable. I will keep it in mind, thank you! $\endgroup$
    – Eduardo M.
    Apr 21 '16 at 2:16
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Sorry I deleted my comment. I understand your question a little bit better now.

Since the two curves are graphs of the functions $f(x) = x^3 -3x+1$ and $g(x) = 9x+17$, they intersect at points $c$ where $f(c) = g(c)$. This is equivalent to saying $f(c) - g(c) = 0$, which is equivalent to saying the function $h(x) = f(x) - g(x)$ has a zero at $c$.

If the curves are tangent at a point of intersection, then not only $f(c) = g(c)$ is true but $f'(c) = g'(c)$ as well. That is, $f$ and $g$ share the same tangent line (point and slope) at $c$. The equation $f'(c) = g'(c)$ is equivalent to $h'(c) = 0$. So we need to solve the equations $h(x) = 0$ and $h'(x) = 0$.

One way to solve the problem (my original suggestion) is to find the zeroes of $h$ and plug them into $h'$ to see if they are also zeroes of $h'$. The trouble is that $h$ is cubic so finding roots is a hair more difficult than standard high school algebra. But $h'$ is quadratic so you can find its zeroes easily. You found them: they are $\pm 2$. But only one of these ($-2$) is also a root of $h$. So the point of tangency is $(-2,-1)$.

The IVT is redundant, you don't need to otherwise prove that the function has a zero if you can find it explicitly.

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  • $\begingroup$ Thank you! This is exactly what I was failing to see, your answer was really helpful. And I understand now that the first use of Bolzano's Theorem is redundant. $\endgroup$
    – Eduardo M.
    Apr 21 '16 at 1:41

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