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C H A P T E R 2 Polynomial and Rational Functions

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256 Chapter 2 <strong>Polynomial</strong> <strong>and</strong> <strong>Rational</strong> <strong>Functions</strong><br />

137. (a)<br />

y<br />

2 in. 2 in.<br />

(b) The area of print is x 4y 4, which is<br />

30 square inches.<br />

0<br />

0<br />

x 4y 4 30<br />

x<br />

2 in.<br />

2 in.<br />

y 4 30<br />

x 4<br />

y 30<br />

4<br />

x 4<br />

y <br />

y <br />

y <br />

30 4x 4<br />

x 4<br />

4x 14<br />

x 4<br />

22x 7<br />

x 4<br />

22x 7 2x2x 7<br />

Total area xy x x 4 <br />

x 4<br />

100<br />

(c) Because the horizontal margins total 4 inches, x must be<br />

greater than 4 inches. The domain is x > 4.<br />

(d)<br />

200<br />

4 32<br />

0<br />

The minimum area occurs when x 9.477 inches, so<br />

y <br />

22 9.477 7<br />

9.477 4<br />

9.477 inches.<br />

The least amount of paper used is for a page size of about<br />

9.48 inches by 9.48 inches.<br />

138.<br />

The limiting amount of uptake is determined<br />

by the horizontal asymptote,<br />

y <br />

90<br />

18.47<br />

0.23 80.3 mgdm2 18.47x 2.96<br />

y , 0 < x<br />

0.23x 1<br />

139.<br />

CO2 hr.<br />

Critical numbers:<br />

Test intervals: , <br />

Test: Is 3x 42x 1 < 0?<br />

4<br />

x <br />

3 , , 43<br />

4<br />

6x<br />

3x 42x 1 < 0<br />

1<br />

3 , x 2<br />

2 6x<br />

5x 4 < 0<br />

2 5x < 4<br />

140.<br />

2 x 2 x ≥ 15 141.<br />

2 x 2 x 15 ≥ 0<br />

Critical numbers: x <br />

Test intervals: , 3 ⇒ 2x 5x 3 > 0<br />

5<br />

2x 5x 3 ≥ 0<br />

2 , x 3<br />

5 2 , 3,<br />

⇒ 2x 5x 3 > 0<br />

5<br />

2 ⇒ 2x 5x 3 < 0<br />

Solution interval: , 3 5<br />

2 , <br />

1<br />

, 2, 1 2 , <br />

By testing an x-value in each test interval in the<br />

inequality, we see that the solution set is: 4<br />

3<br />

x 3 16x ≥ 0<br />

xx 4x 4 ≥ 0<br />

Critical numbers:<br />

Test intervals:<br />

x 0, x ±4<br />

Test: Is xx 4x 4 ≥ 0?<br />

By testing an x- value in each test interval in the inequality,<br />

we see that the solution set is: 4, 0 4, .<br />

, 1<br />

2<br />

, 4, 4, 0, 0, 4, 4,

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