Homework SoursePn12 approximates f12 to within 0005SolutionThe tangent line
Pn(1/2) approximates f(1/2) to within 0.005
Solution
The tangent line to the graph of y = f(x) at the point x = a is the line going through the point ( ) a, f (a) that has slope f \'(a) . By the point-slope form of the equation of a line, its equation is y f (a) = f \'(a)(x a) y = f (a) + f \'(a)(x a)
We will give a name, T1(x) , to the function corresponding to the tangent line: T1(x) = f (a) + f \'(a)(x a) For x near x = a, we have f (x) T1(x). The tangent line function T1(x) is called the Taylor polynomial of degree one for f(x), centered at x = a. Notice that it satisfies the two conditions T1(a) = f (a) and T1 \'(a) = f \'(a). In other words, T1(x) is the polynomial of degree one that has the same function value at x = a and the same first derivative value at x = a as the original functions f(x). We can get a better approximation, T2 (x) near x = a, using a parabola (a polynomial of degree two). The formula for T2 (x) is T2 (x) = f (a) + f \'(a)(x a) + f \'\'(a) 2 ( ) x a 2 . T2 (x) is called the Taylor polynomial of degree two for f(x), centered at x = a. SinceT2 \'(x) = f \'(a) + f \'\'(a)(x a) and T2 \'\'(x) = f \'\'(a) , T2 (x) satisfies the three conditions T2 (a) = f (a), T \' 2 (a) = f \'(a) and T2 \'\'(a) = f \'\'(a). In other words, T2 (x) is the polynomial of degree two that has the same function value at x = a, the same first derivative value at x = a, and the same second derivative value at x = a as the original function f(x).
