Answer
$3+\frac{1}{6}x-\frac{1}{216}x^{2}+\frac{1}{1296}x^{3}-\frac{5}{23328}x^{4}$
Work Step by Step
$f(x)=\sqrt x+9$
$f(x)=( x+9)^{\frac{1}{2}}$
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
$P_{4}(x)=f(0)+\frac{f^{(1)}(0)}{1!}x+\frac{f^{(2)}(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}+\frac{f^{(4)}(0)}{4!}x^{4}$
$=3+\frac{1}{6}x-\frac{1}{216}x^{2}+\frac{1}{1296}x^{3}-\frac{5}{23328}x^{4}$