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This is a didactic question.

Given a differentiable function $y=f(x) \;, x,y \in \mathbb{R}$, I want to construct an exercise in which we have to find a straight line that passes through a point $P=(x_P,y_P)=(x_P, f(x_P))$ of its graph and is tangent in some other point $X=(x,y)=(x,f(x))$.

This problem reduced to solve the equation: $$ (1) \qquad f(x)-f(x_P)=f'(x)(x-x_P) $$ that has the obvious solution $x=x_P$, but the other solutions (if they exists) are not, in general, ''easy'' solutions , i.e. solutions that can be expressed in closed form with elementary functions.

My problem is to find some class of functions such that equation $(1)$ can have such ''easy'' solutions.

I've find that functions of the form $$ f(x)=ax+\dfrac{b}{x^2}+c $$ work well, since equation $(1)$ in this case become a second degree equation that can be solved in elementary way. But I'm searching for some other class of functions and since I don't have a method to look for them, my imagination failed to produce any interesting result.

So my question is: someone can find a method ( but I doubt that exists) or has enough imagination to find some other class of functions tha work well for this kind of exercises?

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    $\begingroup$ Choose any polynomial $f(x)$ nice enough so that you can factor the roots of the resulting polynomial equation. $\endgroup$
    – abnry
    Mar 21 '15 at 19:58
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    $\begingroup$ Expanding on nayrb's comment, it's a pleasant exercise that if $f$ is a quadratic polynomial, there exist two distinct (real) tangents through $P$ if and only if $P$ lies "on the non-convex side" of the graph of $f$ (e.g., below the graph $y = x^{2}$). $\endgroup$ Mar 21 '15 at 20:05
  • $\begingroup$ Thanks @nayrb. This is a way that I have tried, but it's not easy to find polynomials of degree $\ge 3$ for which the equation can be easily factored. $\endgroup$ Mar 21 '15 at 20:05
  • $\begingroup$ Obviously any polynomial of odd degree has some tangent that intercept his graph in some other point, but the problem is to construct an exercise in which the tangent can be find in an elementary way strating from this point. $\endgroup$ Mar 21 '15 at 20:14
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Let's consider a cubic function $f(x)=a_0x^3+a_1x^2+a_2x+a_3$. Assume that you know $(p,q)$ a point on the graph of the function. Then, you know that $q=a_0p^3+a_1p^2+a_2p+a_3$. Using this, you can find a relation for $a_3=q-a_0p^3-a_1p^2-a_2p$. So, we write $a_3$ as a polynomial in $a_0$, $a_1$, and $a_2$.

Now, consider the line $y-q=m(x-p)$. This is a line passing through the point $(p,q)$. We would like this line to be tangent to the curve at another point. Then, consider the line as $y=mx-mp+q$ and by substitution, we consider $$ mx-mp+q=a_0x^3+a_1x^2+a_2x+a_3. $$ A solution to this equation is both on the graph of the cubic function and on the line above. Consider $$ a_0x^3+a_1x^2+(a_2-m)x+(a_3+mp-q)=0. $$ This is a cubic polynomial in $x$ and we are interested in its roots. We know that $x=p$ is a root by the definition of $a_3$. So, we can rewrite this as $$ (x-p)(a_0p^2+pa_1+pa_0x-m+a_2+xa_1+a_0x^2)=0. $$ (I used maple to get this factorization and to eliminate $a_3$).

Now, to be a tangent point to the curve, the remaining factor must have a double root. You can observe that this has a double root when the discriminant vanishes. Therefore, you want $$ (a_0p+a_1)^2-4a_0(a_0p^2+pa_1-m+a_2)=0 $$

Therefore, you should choose $a_0$, $a_1$, $a_2$, and $p$ so that this equation holds. This will generate the type of curve that you're looking for.

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  • $\begingroup$ Good answer (upvote). It seems that from the discriminant we can always find $m$ for any $a_0,a_1,a_2,p$ so that this parameters can be chosen in such away that also $f(x)$ can be studied with elementary methods. But the function is not so different from mine (that is a fraction with a cubic numerator). We'll see if someone has some totally different solution. $\endgroup$ Mar 21 '15 at 21:45
  • $\begingroup$ @EmilioNovati Note that this construction can be extended to any degree polynomial. You just need to check that the appropriate discriminant vanishes (which is much harder to compute in general). $\endgroup$ Mar 21 '15 at 21:53
  • $\begingroup$ Yes! For even degree $n>3$ we can request that also $x=p$ is a double root and this gives the problem to find ''multi-tangents'', that is another interesting exercise, at least in my opinion. I've posted a question about this but without success math.stackexchange.com/questions/1076688/… . $\endgroup$ Mar 21 '15 at 22:06
  • $\begingroup$ @EmilioNovati Your other question is also interesting. I can use the same tricks as above to get the bi-tangent case, and I have some ideas for the tri-tangent case. $\endgroup$ Mar 21 '15 at 22:49
  • $\begingroup$ @EmilioNovati I found a very nice and easy solution. See the question that you linked above. $\endgroup$ Mar 22 '15 at 0:40

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