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

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

Answer 2

Comparing with the general Taylor expansion, we will see that the coefficients are:

• f(0) = 4
• f'(0) = -3
• f''(0) = 2!*5 = 10
• f'''(0) = 3!*4 = 3*2*4 = 24

How to work with a Taylor polyomial?

For a function f(x), the Taylor polynomial around x = 0 is:

T(x) = f(0) + f'(0)*(x) + (1/2!)*f''(0)*x^2 + (1/3!)*f'''(0)*x^3 + ...

In this case, we have:

T(x)  = 4 − 3x + 5*x^2  + 4*x^3

Comparing the correspondent coefficients with the ones in the general expansion, we can see that:

• f(0) = 4
• f'(0) = -3
• f''(0) = 2!*5 = 10
• f'''(0) = 3!*4 = 3*2*4 = 24

If you want to learn more about Taylor expansions, you can read:

https://brainly.com/question/9211177