But if you have a Texas Instruments TI 84 (or similar) calculator, then you can do the following:

In the $y_1=$ menu, enter the equation for which you are trying to find the slope of the tangent. Here, $y=x^{-1}$ has been entered (which is the same as $y=\frac{1}{x}$).

The second equation ($y_2=$) finds the slope of a secant to the curve in $y_1$ between $x_1=0.5$ and $x_2=x$. By pasting the function $y_1$ into the slope formula in $y_2$ we can use the same $y_2$ function for any $y_1$! This means that $y_2$ gives the slope of a secant between $x_1=0.5$ and $x_2=x$ no matter what $y_1$ is! This can save a lot of work if you need to find the slope of several tangent of several different functions!

If this sounds confusing, then notice that $y_2$ just gives the formula for the slope between two points, ${m=\frac{f(x_1)-f(x_2)}{x_1-x_2}}$ but where the values in the numerator (the $y$ values) are confined to the rule defined by $y_1$.

This is exactly what we did in the applet above. To make this more clear, let us graph the above two functions.

At the right we see the two functions graphed. The horizontal axis goes from $0$ to $1$ and the vertical axis goes from approximately $-6$ to $4$ or $5$. $y_1$ is above the $x$-axis and $y_2$ (the slope beteen $x=0.5$ and $x$) is below the $x$-axis.

Note that the slope of $y_1$ is always negative. Accordingly, $y_2$ always has negative values.

On $y_2$ we trace to $x=1$. The $y$ value here is $y=-2$. This is the slope of the secant crossing $y_1$ at $x_1 = 0.5$ and $x_2 = 1$ at the very edge of the graph.

This is exactly what we got in our previous calculations.

On $y_2$ we trace to $x=.59574468$. The $y$ value here is $y=-3.357143$. This is the slope of the secant crossing $y_1$ at $x_1 = 0.5$ and $x_2 =.59574468$. Since we have moved our point $x_2$ closer to $x_1$, we can expect that this is a closer approximation to the slope of the tangent at $x_1 = 0.5$ than the slope of the previous secant.

In this graph we have found the slope of the secant between $x_1 = 0.5$ and $x_2 = 0.5001$. This secant is getting very close to the tangent at $x_1=0.5$. The slope of this secant is $m=-3.9992$. The slope of the tangent is probably very close to this value.

Let us confirm our hunch by moving $x_2$ to the other side of $x_1$ but still very close to $x_1$ , namely to $0.499$. We now have the slope of the secant as $-4.008016$.

From the above calculations, it seems that the slope of the tangent of $\,f(x)=x^{-1}$ at $(0.5,2)$ is probably $-4$. (The same as we got before but now using our graphing calculator).

If we succumb to the temptation to trace $x_2$ to $0.5$, then here is the result: there is no $y$ value given because it is undefined. Notice the small “hole” in the graph at that point. We end up trying to divide $0$ by $0$ and that just does not work! We seem unable to get the slope of a tangent without taking an intelligent guess. We get so close, and then boom! we lose it!

Nevertheless, we can be quite sure that the slope of the tangent to $\,f(x)=x^{-1}$ at $(0.5,2)$ is equal to $-4$. Using that value and knowing a point is goes through, we can get the equation of the tangent.

You can do the math yourself. The equation of the tangent line ends up being $y=-4x+4$. We have entered that equation into $y_3$ in our TI 84.

Lastly, just so we can see that we've had a successful result, we can see the graph of the tangent on our calculator.

Though we have been successful, the method we have used is cumbersome and lacks precision. We are going to be on the lookout for a better method!