Answer to Question #251506 in Calculus for Matt

Question

Expand f(x) = $e^{2x} + 3x^2$ in a Taylor series at a = 3 to the third derivative degree

Answer 1

Solution:

$f(x)=e^{2x}+3x^2 \\f'(x)=2e^{2x}+6x \\f''(x)=4e^{2x}+6 \\f'''(x)=8e^{2x}$

Then, at $x=3$,

$f(3)=e^6+27 \\f'(3)=2e^6+18 \\f''(3)=4e^6+6 \\f'''(3)=8e^6$

We know that, Taylor series expansion at a = 3 to the third derivative degree is

$f(x)=P_3(x)=f(3)+f'(3)(x-3)+\dfrac{f''(3)}{2}(x-3)^2+\dfrac{f'''(3)}{3\times2}(x-3)^3$

$=e^6+27+(2e^6+18)(x-3)+\dfrac{4e^6+6}2(x-3)^2+\dfrac{8e^6}6(x-3)^3$

$=\dfrac{4e^6x^3-30e^6x^2+9x^2+78e^6x-69e^6}{3}$