## Calculus: Early Transcendentals 8th Edition

$f(x) = x + 2x^{\frac{3}{2}} + 5$
$f'(x) = 1 + 3x^{\frac{1}{2}}$ $f(x) = \int 1 + 3x^{\frac{1}{2}}dx$ $f(x) = x + \frac{3x^{\frac{1}{2}}}{\frac{3}{2}} + C$ $f(x) = x + 2x^{\frac{3}{2}} + C$ Given $f(4) = 25$: $f(4) = 4 + 2(4)^{\frac{3}{2}} + C$ $f(4) = 4 + 2(8) + C$ $f(4) = 4 + 16 + C$ $25 = 20 + C$ $25 - 20 = C$ $5 = C$ Substitute: $f(x) = x + 2x^{\frac{3}{2}} + 5$