Answer
(a) $C'(x) = 3+0.02x+0.0006x^2$
(b) $C'(100) = 11$
This value tells us the cost of producing each additional item when the production level is at 100. It gives us an estimate that the cost of producing the 101st item will be approximately $\$11$
(c) The cost of producing the 101st item is $\$11.07$ which is just 7 cents more than $C'(100) = \$11$
Work Step by Step
(a) $C(x) = 2000+3x+0.01x^2+0.0002x^3$
$C'(x) = 3+0.02x+0.0006x^2$
(b) $C'(x) = 3+0.02x+0.0006x^2$
$C'(100) = 3+0.02(100)+0.0006(100)^2$
$C'(100) = 3+2+6$
$C'(100) = 11$
This value tells us the cost of producing each additional item when the production level is at 100. It gives us an estimate that the cost of producing the 101st item will be approximately $\$11$
(c) $C(101) = 2000+3(101)+0.01(101)^2+0.0002(101)^3$
$C(101) = \$2611.07$
$C(100) = 2000+3(100)+0.01(100)^2+0.0002(100)^3$
$C(100) = \$2600$
$C(101)-C(100) = \$11.07$
The cost of producing the 101st item is $\$11.07$ which is just 7 cents more than $C'(100) = \$11$
