We digressed from our "difference between x3 and x2" discussion when someone asked, "Can we get GC to draw their tangents?" This is a reprise of what we did subsequently:

1. The function r(x) = (f(x+h) - f(x))/h, h>0, gives an approximation to how fast a function is changing for each value of x.
2. The tangent to f at x = a passes through the point (a,f(a)) with rate r(a). So the function t(u) = r(a)u + b will produce the tangent to f at x = a. We only need to calculate the value of b when x = a.
3. Since r(a) is the rate of change of t when x = a, whenever x changes by δx, t will change by r(ax. So, to find the value of t(0), we need to change x by -a. The value of t will change by r(a)(-a). But the value of t(a) is f(a). So t(0) = f(a) + r(a)(-a).
4. The tangent to f at x=a will be t(x) = r(a)x + (f(a) - r(a)(-a)).

The following movie is what we saw when we defined t as above, defined a by way of a slider whose values range from -10 to 10, and we let h = 0.01, and we defined f as f(x) = x3. (Click here for the original GC file.)

Then we thought to see how general our solution was by changing the definition of f to f(x)=cos(10sin(x)). What we saw is shown in the movie below. (Click here for the GC file.)