Answer to Question #335067 in Calculus for Johannes Steven

Question

Find the Maclaurin series for the function f(x) = square root of

(x²+2-x)⁵

Answer 1

$p_0(0)=f(0)=\sqrt{2^5}=\sqrt{32}$

$f'(x)=\frac{5(2x-1)(x^2-x+2)^{3/2}}{2}$

$p_1(0)=f(0)+f'(0)x=\sqrt{32}-\frac{5\sqrt{8}}{2}x$

$f''(x)=5(x^2-x+2)^{3/2}+\frac{15(2x-1)^2\sqrt{x^2-x+2}}{4}$

$p_2=\sqrt{32}-\frac{5\sqrt{8}}{2}x+ \frac{5\sqrt{8}}{2}x^2+\frac{15\sqrt{2}}{8}x^2$