Question

calculator required
example 6: the table gives the life expectancy of us females born in a given year over successive 50-year intervals. over which of the following intervals is the average rate of change in life expectancy the greatest?
(a) from 1800 to 1850
(b) from 1850 to 1900
(c) from 1900 to 1950
(d) from 1950 to 2000

birthyear	1800	1850	1900	1950	2000

Explanation:

The average rate of change of a function \( f(x) \) over the interval \([a, b]\) is given by the formula:
\[
\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}
\]
For each interval, we calculate the average rate of change using the life expectancy values ( \( f(x) \) ) and the time interval ( \( b - a = 50 \) years, since each interval is 50 years).

Step 1: Interval (A): 1800 to 1850

• \( a = 1800 \), \( f(a) = 41.24 \)
• \( b = 1850 \), \( f(b) = 46.10 \)
• Average Rate of Change:

\[
\frac{46.10 - 41.24}{1850 - 1800} = \frac{4.86}{50} = 0.0972
\]

Step 2: Interval (B): 1850 to 1900

• \( a = 1850 \), \( f(a) = 46.10 \)
• \( b = 1900 \), \( f(b) = 53.63 \)
• Average Rate of Change:

\[
\frac{53.63 - 46.10}{1900 - 1850} = \frac{7.53}{50} = 0.1506
\]

Step 3: Interval (C): 1900 to 1950

• \( a = 1900 \), \( f(a) = 53.63 \)
• \( b = 1950 \), \( f(b) = 70.65 \)
• Average Rate of Change:

\[
\frac{70.65 - 53.63}{1950 - 1900} = \frac{17.02}{50} = 0.3404
\]

Step 4: Interval (D): 1950 to 2000

• \( a = 1950 \), \( f(a) = 70.65 \)
• \( b = 2000 \), \( f(b) = 81.83 \)
• Average Rate of Change:

\[
\frac{81.83 - 70.65}{2000 - 1950} = \frac{11.18}{50} = 0.2236
\]

(Revised):
The average rate of change of a function \( f(x) \) over \([a, b]\) is \( \frac{f(b) - f(a)}{b - a} \). For each 50 - year interval (\( b - a = 50 \)):

• Interval (A) (1800–1850):

\( \frac{46.10 - 41.24}{50} = \frac{4.86}{50} = 0.0972 \)

• Interval (B) (1850–1900):

\( \frac{53.63 - 46.10}{50} = \frac{7.53}{50} = 0.1506 \)

• Interval (C) (1900–1950):

\( \frac{70.65 - 53.63}{50} = \frac{17.02}{50} = 0.3404 \)

• Interval (D) (1950–2000):

\( \frac{81.83 - 70.65}{50} = \frac{11.18}{50} = 0.2236 \)

Comparing the rates: \( 0.3404 > 0.2236 > 0.1506 > 0.0972 \). Thus, the greatest average rate of change is in interval (C).

Answer:

(D) from 1950 to 2000 (Wait, no—wait, in Step 3, we calculated 0.3404, which is higher than Step 4’s 0.2236. Wait, I made a mistake. Let’s recheck:

Wait, 70.65 - 53.63 = 17.02, divided by 50 is 0.3404. Then 81.83 - 70.65 = 11.18, divided by 50 is 0.2236. Wait, so actually, the greatest is (C)? Wait, no—wait, the options are (A) 0.0972, (B) 0.1506, (C) 0.3404, (D) 0.2236. So (C) is higher? Wait, but let me recheck the numbers:

Life Expectancy:
1800: 41.24
1850: 46.10 (diff: 4.86)
1900: 53.63 (diff from 1850: 7.53)
1950: 70.65 (diff from 1900: 17.02)
2000: 81.83 (diff from 1950: 11.18)

Ah, so 1900 to 1950 has a larger difference. Wait, but the original problem—maybe I misread the options. Wait, the options are (A) 1800-1850, (B) 1850-1900, (C) 1900-1950, (D) 1950-2000.

So recalculating:

(A): (46.10 - 41.24)/50 = 4.86/50 = 0.0972

(B): (53.63 - 46.10)/50 = 7.53/50 = 0.1506

(C): (70.65 - 53.63)/50 = 17.02/50 = 0.3404

(D): (81.83 - 70.65)/50 = 11.18/50 = 0.2236

So the largest average rate of change is (C) from 1900 to 1950? Wait, but maybe I made a mistake. Wait, the problem says "the greatest"—so (C) is 0.3404, which is higher than (D)’s 0.2236. So the correct answer is (C)? Wait, but let me check again.

Wait, 70.65 - 53.63 is 17.02, divided by 50 is 0.3404. 81.83 - 70.65 is 11.18, divided by 50 is 0.2236. So yes, (C) has a higher rate. So my initial mistake was in the final answer. Let’s correct that.