## Calculus with Applications (10th Edition)

$3+\frac{1}{6}x-\frac{1}{216}x^{2}+\frac{1}{1296}x^{3}-\frac{5}{23328}x^{4}$
$f(x)=\sqrt x+9$ $f(x)=( x+9)^{\frac{1}{2}}$ Find the Taylor polynomials of degree 4 at 0. then $n=4$ The $n - th$ derivatives are $f^{(1)}(x)=\frac{d}{dx}[( x+9)^{\frac{1}{2}}]=\frac{1}{2}( x+9)^{\frac{-1}{2}}$ $f^{(2)}(x)=\frac{-1}{4}( x+9)^{\frac{-3}{2}}$ $f^{(3)}(x)=\frac{3}{8}( x+9)^{\frac{-5}{2}}$ $f^{(4)}(x)=\frac{-15}{16}( x+9)^{\frac{-7}{2}}$ Evaluate $f(0),f^{(1)}(0),f^{(2)}(0),f^{(3)}(0),f^{(4)}(0)$ $f^{(0)}(0)=( 0+9)^{\frac{1}{2}}=3$ $f^{(1)}(0)=\frac{1}{2}(0+9)^{\frac{-1}{2}}=\frac{1}{6}$ $f^{(2)}(0)=\frac{-1}{4}(0+9)^{\frac{-3}{2}}=\frac{-1}{108}$ $f^{(3)}(0)=\frac{3}{8}(0+9)^{\frac{-5}{2}}=\frac{1}{648}$ $f^{(4)}(0)=\frac{-15}{16}(0+9)^{\frac{-7}{2}}=\frac{-5}{11664}$ Replace the found values into the definition of Taylor Polynomial for n=4 $P_{4}(x)=f(0)+\frac{f^{(1)}(0)}{1!}x+\frac{f^{(2)}(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}+\frac{f^{(4)}(0)}{4!}x^{4}$ $=3+\frac{1}{6}x-\frac{1}{216}x^{2}+\frac{1}{1296}x^{3}-\frac{5}{23328}x^{4}$