Answer
(a) $2011: 42.48$ million people
$2030: 70.87$ million people
(b) 2011: 1.18 million people per year
2030: 1.51 million people per year
(c) $2.13\%$ per year
Work Step by Step
$n(x)=-0.00082x^3+0.059x^2+0.183x+34.42$
Putting $x=11$
$n(11)=-0.00082(11^3)+0.059(11^2)+0.183(11)+34.42=42.48$ million people
(Correct to two significant digits)
Putting $x=30$
$n(30)=-0.00082(30^3)+0.059(30^2)+0.183(30)+34.42=70.87$ millions people
Hence
2011: 42.48 million people
2030: 70.87 million people
(b)
Taking derivative with respect to $x$ of the expression
$n(x)=-0.00082x^3+0.059x^2+0.183x+34.42$
$n^{'}(x)=-0.00082(3x^2)+0.059(2x)+0.183$
Put $x=11$
$n^{'}(11)=-0.00082[3(11^2)]+0.059(2\times 11)+0.183=1.18$ million people per year
Put $x=30$
$n^{'}(x)=-0.00082[3(30^2)]+0.059(2\times 30)+0.183=1.51$million people per year
(Correct to two significant digits)
Hence
2011: 1.18 million people per year
2030: 1.51 million people per year
(
c)
Percentage rate of change in 2030=[ ( rate of change in 2030)/
(value of the function in 2030) ] $ \times 100$
rate of change in 2030=1.51
value of the function in 2030=70.87
Percentage rate of change in 2030=$\frac{1.51}{ 70.87 }\times 100= 2.13\%$
