Solve the problem.
-The population of a certain country since 1990 can be approximated by $f ( t ) = 0.008 t ^ { 2 } + 2.1 t + 175$ where $f ( t )$ is the population in millions and $t$ represents the number of years since 1990 . Find the average rate of change in the country's population between 1990 and 2010 . Round to 1 decimal place.
A) $2.3$ million/yr
B) $2.0$ million/yr
C) $2.5$ million $/ \mathrm { yr }$
D) $2.9 \mathrm { million } / \mathrm { yr }$

Question

Quizplus · Accepted Answer

The average rate of change of a function over an interval $[a, b]$ is given by $\frac{f(b) - f(a)}{b - a}$. Here, $a = 0$ (1990) and $b = 20$ (2010), so we calculate $\frac{f(20) - f(0)}{20 - 0}$. Substituting into the given function, we get $\frac{(0.008(20)^2 + 2.1(20) + 175) - (0.008(0)^2 + 2.1(0) + 175)}{20}$. Simplifying, we find $\frac{(0.008(400) + 42 + 175) - 175}{20} = \frac{3.2 + 42}{20} = \frac{45.2}{20} = 2.26$, which rounds to $2.3$ million per year.

Solve the Problem f(t)=0.008t2+2.1t+175f ( t ) = 0.008 t ^ { 2 } + 2.1 t + 175f(t)=0.008t2+2.1t+175

Solve the Problem $f ( t ) = 0.008 t ^ { 2 } + 2.1 t + 175$