# Given #f(x)= x^2 + 3x + 8# how do you determine the first three Taylor polynomials at x=0?

##### 1 Answer

#f(x)= x^2 + 3x + 8#

#### Explanation:

A Taylor Series about

They both use a polynomials to estimate the function. The more terms you add, the closer the polynomial becomes to the function

A polynomial is therefore its own series

Hence the first three terms (and the only terms) of the series for

ie

#f(x)= x^2 + 3x + 8#

**Proof**

If you are not convinced by the above explanation we can see by direct proof using the definition of the Maclaurin series (The TS about

By definition:

# f(x) = f(0) + f'(0)x + (f''(0))/(2!)x^2 + (f^((3))(0))/(3!)x^3 + ... + (f^((n))(0))/(n!)x^n + ... #

We have:

# \ \ \ \ \ f(x) = x^2 + 3x + 8 #

# :. f(0) = 8 #

Differentiate wrt to get the first derivative:

# \ \ \ \ \ f'(x) = 2x+3 #

# :. f'(x) = 3 #

Differentiate again wrt to get the second derivative:

# \ \ \ \ \ f''(x) = 2 #

# :. f''(0) = 2 #

Differentiate again wrt to get the third derivative:

# \ \ \ \ \ f^((3))(x) = 0 #

All further derivatives are zero

So we can now form the Maclaurin series:

# f(x) = 8 + 3x + 2x^2/2 + 0 x^3 + 0x^4 +... #

# \ \ \ \ \ \ \ = 8 + 3x + x^2 #

Which is exactly the same polynomial that we started with.