Answer
$C(x)=10x+\displaystyle \frac{x^{2}}{2}-\frac{1}{x}+4000.01$
Work Step by Step
The marginal cost at production level of x is
$C'(x)=10+x+\displaystyle \frac{1}{x^{2}}=10+x+x^{-2}$
So, $C(x)$ is an antiderivative:
$C(x)=\displaystyle \int(10+x+x^{-2})dx$
$=10x+\displaystyle \frac{x^{2}}{2}+\frac{x^{-1}}{-1}+D$
$=10x+\displaystyle \frac{x^{2}}{2}-\frac{1}{x}+D$
The indefinite integral gives us a collection of functions.
To find the exact function, we find $D.$
The text gives us:$\quad C(100)=10,000$,
from which we find $D.$
$10,000=10(100) +\displaystyle \frac{100^{2}}{2}-\frac{1}{100}+D$
$10,000=1000+5000-0.01+D$
$4000.01=D$
Thus,
$C(x)=10x+\displaystyle \frac{x^{2}}{2}-\frac{1}{x}+4000.01$
