Tangent of a rational function

Question

I'm wondering if someone could help me solve a little conundrum I've been having. I'm trying to find the equation of a line with slope $-1$ tangent to the curve of a rational function. The function is $y = 1/(x-1)$ and the general equation of the line would be $y = -x+k$. I understand that if I set the $y$-values equal -- that is, $1 / (x-1)= -x + k$ -- I should be able to reorder it into a quadratic and solve for k, giving me the equation of the line. So far, the quadratic I'm finding is $y = -x^2+x+kx-k-1$ , yet I can't seem to find a satisfactory value for k that would give me a tangent to the curve. Any thoughts?

Answer 1

Bad news is, you can't really do anything with this without calculus.

To find a thing tangent to the curve you have to, of course, take the derivative. Derivative of this looks like (use the chain rule with $u=x-1$)

$$\frac{dy}{dx} = -\frac{1}{\left(x-1\right)^2}$$

SInce we're looking for a place where the derivative has value $-1$, we get

$$-\frac{1}{\left(x-1\right)^2}=-1$$

Which is easily solvable:

$$\left(x-1\right)^2=1$$

$$x^2-2x = 0$$

$$x \in \left\{0, 2\right\}$$

We can then use these two $x$ values to find $y$ values on the original function; the two tangent lines (and there are two) each go through one of these points.

Answer 2

The approach you're following in your question is correct, except that last equation should have $0$ instead of $y$ i.e. the quadratic equation $-x^2+(k+1)x-(k+1)=0$. The quadratic formula immediately yields roots $$x=-\frac{1}{2}\left(-(k+1)\pm \sqrt{(k+1)^2-4(k+1)}\right)=\frac{1}{2}\left(k+1\pm \sqrt{k^2-2k-3}\right)$$

Note that the type of roots is determined by the sign of discriminant (i.e. the argument of the square root) and we have two coinciding roots when it vanishes. So we solve $k^2-2k-3=0$ for $k=-1,3$ by inspection. From this we deduce three cases:

• $-1<k<3$: The discriminant is negative and so there are no real roots
• $k>3$ or $k<-1$: The discriminant is positive and so there are two real roots
• $k=-1,3$: The discriminant is zero, so there is exactly one real root

All that remains is to interpret these three cases geometrically...

Answer 3

A more visual approach that might be algebraically simpler -- and more in the spirit of the precalculus tag -- (but does not generalize to harder problems) is to realize that $y=1/(x-1)$ is a translation one unit to the right of $y=1/x$. So let's solve the simpler problem of finding a line of the form $y=-x+k$ that is tangent to $y=1/x$; then we just have to shift the line one unit to the right (which is the same as shifting it one unit up).

Take a look at the graph below. Because the graph has such nice symmetries you can see very easily that the line will be tangent to the curve only at the points $(1,1)$ and $(-1,-1)$, and it is easy to see what values of $k$ will produce those two intersections. So the simpler problem is solved. Now to solve your original problem just translate everything back one unit to the right.