Question

3. Suppose that a function f(x) is approximated near a = 0 by the 3rd degree Taylor polynomial T3(x) = 4 − 3x + x 2 5 + 4x 3 . Give the values of f(0), f0 (0), f00(0), and f 000(0).

Answer 1

Answer:

$f^{(0)}(0)} = 4 \\\\\ f^{(1)}(0) = -3 \\\\\f^{(2)}(0) = 4 \\\ \\f^{(3)}(0) = 4*3! = 24$

Step-by-step explanation:

Remember that in general  the taylor series of a function   $f$   are given by

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = \frac{f^{0}(0)}{0!}+\frac{f^{(1)}(0)}{1!} x+ \frac{f^{(2)}(0)}{2!} x^2 + \frac{f^{(3)}(0)}{3!} x^3 + ......$

therefore

$\frac{f^{(0)}(0)}{0!} = 4 \\\frac{f^{(1)}(0)}{1!} = -3 \\\\\frac{f^{(2)}(0)}{2!} = 1\\\frac{f^{(3)}(0)}{3!} = 4$

and then

$f^{(0)}(0)} = 4 \\\\\ f^{(1)}(0) = -3 \\\\\ f^{(2)}(0) = 4 \\\ \\f^{(3)}(0) = 4*3! = 24$