# 7.1. Constructing Tangents

In this chapter we are trying to create a tangent to a given function  f. Intuitively a tangent is a straight line touching the graph of  f at a fixed point (a, f(a)) snuggling as close as possible.

Amongst the infinitely many linear functions, i.e. functions of the $mX+b$ type that pass (a, f(a)) we are looking for the one having the 'right' gradient m. Taking  $f≔\frac{1}{4}{X}^{2}$  as an example and choosing $a=2$ we would expect the following scene:

Though the sketch suggests a gradient value of 1, it is still up to us to confirm this by calculation. Normally there is no problem to compute the slope of a straight line: With two different points just take the ratio

$m=\frac{\text{increase in height}}{\text{increase in width}}$ .

Tangents however provide sure access to only one point, namely (2, f(2)), so there is no way to calculate the gradient this simply.

As the slope of the tangent is in some way controlled by shape of  f, we try to succeed with the following detour: We consider all the secants of  f passing (2, f(2)). These are straight lines passing (2, f(2)) and a second point (x, f(x)) .

For these secants the gradient figures ${m}_{2}\left(x\right)$ are easy to get: if $x\ne 2$ the increase in height is the difference of the values of  f  $f\left(x\right)-f\left(2\right)$ - and the increase in width is the difference of the arguments. Thus:

 ${m}_{2}\left(x\right)≔\frac{\frac{1}{4}{x}^{2}-\frac{1}{4}{2}^{2}}{x-2}=\frac{\frac{1}{4}\left(x+2\right)\left(x-2\right)}{x-2}=\frac{1}{4}\left(x+2\right)$ [7.1.1]

The following applet shows the different positions of the secants and calculates their gradients.

We get a double information from this: The closer the second point comes to the fixed one, i.e. the nearer x is to 2, the less differ

• secant and tangent

• ${m}_{2}\left(x\right)$ and the figure 1.

So we may argue: If x converges to 2 the secant gradients ${m}_{2}\left(x\right)$ would approach the tangent gradient 1. That means we are able to calculate the tangent gradient as soon as we could determine the limit $\underset{x\to 2}{\mathrm{lim}}{m}_{2}\left(x\right)$. According to [7.1.1] however, the function ${m}_{2}$ is continuously continuable at 2 by the following limit:

 $\underset{x\to 2}{\mathrm{lim}}{m}_{2}\left(x\right)=\frac{1}{4}\left(2+2\right)=1$. [7.1.2]

Now, this result enables us to construct the tangent, more precisely the tangent function ${t}_{2}$, itself. First $m=1$ provides ${t}_{2}=X+b$ and from the condition ${t}_{2}\left(2\right)=f\left(2\right)⇔2+b=1$ we get $b=-1$. Finally we have

${t}_{2}=X-1$
as the tangent function we looked for. 