Derivatives Without Limits, Part 2

In my last post I talked about how to verify that a point on the graph of a cubic polynomial was a local minimum using only techniques from algebra. We will now try to generalize our methods a bit to be able to confirm our guess of the slope at an arbitrary point on a polynomial graph.

Consider for a moment the function $y=x^2$ at the point $x=1$.



We can guess what the slope of $y=x^2$ at $x=1$ is. By adding the $y=x$ to the graph, we see that the slope needs to be bigger than $1$:

Parabola with wrong tangent line



Adding $y=3x-2$ shows that the slope is less than $3$:

Parabola with wrong tangent line


However, $y=2x-1$ seems to fit just right:

Parabola with correct tangent line


It seems, then, that the slope of $y=x^2$ at $x=1$ is $2$.

So, how can we check if we have the right tangent line? Before, we used the idea of bounce points, although we had to transform our polynomial so that it would bounce off of the $x$-axis. So, we do the same thing here. Algebraically, the simplest transformation that does that is to just subtract $2x-1$ from both functions:


Transformed parabola


Notice that this transformation pulled the point $(1,1)$ down to $(1,0)$. Also, since the parabola was “bouncing” off of the line $y=2x-1$ before the transformation, our new parabola is bouncing off of the $x$-axis at $x=1$.

So, all we have to do is check whether this new parabola, $y=x^2-2x+1$, bounces off of the $x$-axis at $x=1$. Since $x^2-2x+1=(x-1)^2$, we see that this is indeed the case. Therefore, our modified parabola is tangent to the $x$-axis at $x=1$, which means that our original parabola was tangent to $y=2x-1$ at $x=1$.

From this we conclude that the slope of $y=x^2$ at $x=1$ is $2$.

So, our method can actually help us check our guess of the slope of a polynomial at any point, not just extremums. We are not done yet: this method does not even require us to have an initial guess. To see how, read my next post.