Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 240: 91

$f(x)=2x^{3/2}-50$

Step 1. The slope of the tangent line for the curve is given by $f'(x)=3\sqrt x=3(x)^{1/2}$ Step 2. We can find its general antiderivatives as $f(x)=2x^{3/2}+C$ where $C$ is a constant. Step 3. As the function passes through $(9,4)$, we have $2(9)^{3/2}+C=4$ which gives $C=-50$ and the function is $f(x)=2x^{3/2}-50$