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Suppose we have a function f that we find it difficult to
evaluate, but we know a few things about it. We want to make
use of what we know as much as possible to approximate its
value at argument x knowing its value at some argument x0. Given any function f for which we know f(x0) and f '(x0) we can immediately evaluate this approximation. Using it involves pretending that the graph of the function f were its tangent line at x0, rather than whatever it is. We can judge how good this approximation is by considering the second derivative of f. Note that f " is negative in the range between 27 and 28, which means that the first derivative decreases in that interval. On the tangent line at 27 it stays constant. This implies that the tangent line linear approximation at 27 overestimates the change in f over this interval and therefore f(28) < 3 + 1 / 27. We can also observe that f 's third derivative is positive in the interval in question, so that the quadratic approximation, by the same argument underestimates the change in the second derivative and in all lower derivatives as well, so it is a lower bound to f(28), and we get f(28) > 3 + 1 / 27 -1 / (27 * 81). Numerically, to six decimal places we get 3.03658 < f(28) < 3.03704; in fact we have f(28) = 3.03659 to this accuracy. The cube root function considered here is the inverse function to one we know how to calculate, namely the cube. For any such function we can use the linear approximation to obtain it to the accuracy that our machine will allow. |