We digressed from our "difference between *x*^{3} and *x*^{2}" discussion when someone asked, "Can we get GC to draw their tangents?" This is a reprise of what we did subsequently:

- 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*. - 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.* - Since
*r*(*a*) is the rate of change of*t*when*x*=*a*, whenever*x*changes by δ*x*,*t*will change by*r*(*a*)δ*x*. 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*). - 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*) = *x*^{3}. (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.)