Answer
$f\left( x \right) = \frac{{2{x^{3/2}}}}{3} - x - 5$
Work Step by Step
$$\eqalign{
& f'\left( x \right) = \sqrt x - 1,{\text{ }}f\left( 9 \right) = 4 \cr
& f'\left( x \right) = {x^{1/2}} - 1 \cr
& {\text{Find the general antiderivative }}f'\left( x \right) \cr
& {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right) \cr
& {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr
& {\text{Function: }}{x^n}\left( {n \ne - 1} \right) \to {\text{Particular antiderivative: }}\frac{{{x^{n + 1}}}}{{n + 1}} \cr
& {\text{Function: }}\frac{1}{x} \to {\text{Particular antiderivative: }}\ln \left| x \right| \cr
& {\text{we obtain}} \cr
& f\left( x \right) = \frac{{{x^{1/2 + 1}}}}{{1/2 + 1}} - x + {C_1} \cr
& f\left( x \right) = \frac{{{x^{3/2}}}}{{3/2}} - x + {C_1} \cr
& f\left( x \right) = \frac{{2{x^{3/2}}}}{3} - x + {C_1}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Find }}{C_1}{\text{ using the given information }}f\left( 9 \right) = 4 \cr
& 4 = \frac{{2{{\left( 9 \right)}^{3/2}}}}{3} - \left( 9 \right) + {C_1} \cr
& 4 = 18 - 9 + {C_1} \cr
& {C_1} = - 5 \cr
& {\text{Substitute }}{C_1}{\text{ into }}\left( {\bf{1}} \right) \cr
& f\left( x \right) = \frac{{2{x^{3/2}}}}{3} - x - 5 \cr} $$
