Tangent: $ f(p) + f'(p)(x-p) = f(q) + f'(q)(x-q) $
It follows:
$(1)\,$ : $\enspace f'(p) = f'(q)\enspace => \enspace p^3-q^3 = p-q $
$(2)\,$ : $\enspace f(p)-pf'(p) = f(q)-qf'(q)\enspace => \enspace 3(p^4-q^4) = 2(p^2-q^2) $
$(1)$ into $(2)$ leads to $\enspace 3(p^4-q^4) = 2(p+q)(p-q) = 2(p+q)(p^3-q^3)$
which is equivalent to $\enspace (p-q)^2 (p^2-q^2) = 0\,$ .
Because of the condition $\,p\neq q\,$ we get $\,q=-p\,$ .
Putting this result into $(1)$ and choosing $p>0$ we get $p=1$ and therefore $q=-1$ .
The points are $\,(1;f(1))=(1;-2)\,$ and $\,(-1;f(-1))=(-1;0)\,$ .
You can't prove equivalences between a precise, correct mathematical definition, and an imprecise, intuitive, incorrect attempt at a definition.
This idea that the tangent is "the line that only touches the curve once" is, to put it mildly, nonsense. As has been noted, that would mean that every line except the actual tangent is "the tangent" to the horizontal line $x=0$ at any given point. That the vertical line through the origin is the tangent to the $x$-axis at $(0,0)$, etc.
Basically, there are many serious problems with the premises behind your question. But the most important ones:
- It is not true that a line that only touches the graph once near the point is the tangent.
Simply note that for any function $y=f(x)$, the vertical line $x=a$ will only intersect the graph once; yet it almost never is the the tangent line to the graph. And many lines will generally only intersect the graph once: all lines between $y=x+1$ and $y=-x+1$ intersect $y=\cos(x)$ only at $(0,1)$, but none of them are the tangents to $y=\cos(x)$.
- It is not true that the tangent only intersects the graph once near the point.
For example, the function
$$f(x) = \left\{\begin{array}{ll}
x^2\sin\left(\frac{1}{x}\right) &\text{if }x\neq 0,\\
0 &\text{if }x=0
\end{array}\right.$$
has $y=0$ as the tangent at $(0,0)$, but that line intersects the graph infinitely many times in any interval of the form $(-\delta,\delta)$ for $\delta\gt 0$.
Now, you may object that his is a contrived function. It doesn't matter. In fact, there's lots of function, infinitely many of them, for which 2 fails. There is no good notion of "most functions" that you can articulate that will make 2 above true for "most functions". It's just plain wrong.
Now, a better intuition is that the tangent is a straight line that "very closely approximates the graph" at and "near" the point. Can we make that precise?
Yes. Suppose $y=f(x)$ is a graph, and $(a,f(a))$ is a point on the graph. We want a line, $y=mx+b$, with two properties:
$y=mx+b$ goes through the point $(a,f(a))$. For that to happen, we must have $f(a) = ma + b$.
Of all lines that go through the point $(a,f(a))$, the tangent is the one that "best approximates" the graph of $y=f(x)$. That is: the relative error obtained by using the line instead of the function goes to $0$ as $x$ approaches $a$.
The "relative error" is a measure of how big the error is, relative to the size of the input. If I tell you I'm measuring a distance and I'm off by as much as five hundred meters, that's pretty good if I'm trying to figure out how far the Moon is, but it's a pretty lousy approximation if I'm trying to figure out how far the computer screen is from my face.
The absolute error in using the straight line $y=mx+b$ instead of the function $y=f(x)$ would be the distance between the value on the line and the value on the graph of the function. That is, $E(x_0) = |(mx_0+b) - f(x_0)|$.
Because we are trying to be "close to $a$", the relative error is how big this absolute error is, relative to how far we are from $a$. So the relative error at $x_0$ is:
$$R(x_0) = \frac{E(x_0)}{|x-x_0|} = \left|\frac{(mx_0+b)-f(x_0)}{x-x_0}\right|.$$
Definition. The line $y=mx+b$ is tangent to the graph of $y=f(x)$ at the point $(a,f(a))$ if and only if (i) the line goes through the point $(a,f(a))$; and (ii) the relative error goes to $0$ as $x_0$ approaches $a$; that is,
$\lim\limits_{x\to a}R(x) = 0$.
With this definition, it is pretty straightforward to show that the line in question must have slope equal to $f'(a) = \lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$, and so must be the line $y=f'(a)(x-a) + f(a)$.
You can find the derivation and lots of discussion about this issue elsewhere on this site.
Now, you ask, can we prove that the tangent only intersects the graph at the point, at least locally, using the definition?
No. We can't because that's not true in general, as noted above. It's not true that for "most functions" (you would to specify which functions) the tangent line at $(a,f(a))$ will only intersect the graph of $y=f(x)$ at $(a,f(a))$, at least on some interval $(a-\delta,a+\delta)$; it's not true. You can't prove something that is not true, no matter how intuitively you think it ought to be true.
Best Answer
Two curves $y = f(x)$ and $y = g(x)$ have a common tangent line at $x = a$ iff:
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: $$ y - 6 = 9(x - 2) $$