MATH 180 FALL 1995 10:00 SECTION FONG

ASSIGNMENT #9

2.2, #6. f(4) > f(3) since f in increasing.

f(2) − f(1) = \frac{f(2) − f(1)}{2 − 1}, f(3) − f(2) = \frac{f(3) − f(2)}{3 − 2}

are slopes of lines, namely the line connecting (1, (f(1)) and (2, f(2)) and the line connecting (2, (f(2)) and (3, f(3)). Since f is concave-down, f(2) − f(1) > f(3) − f(2). In the same way we see that

\frac{f(2) − f(1)}{2 − 1} > \frac{f(3) − f(1)}{3 − 1}

since the quotients are slopes of lines, namely the line conecting (2, f(2)) and (1, f(1)) and the line connecting (3, f(3)) and (1, f(1)).

2.2, #7. f(4)/4 is the slope of the line connecting (0, 0) and (4, f(4)). Similarly, f(3)/3 is the slope of the line connecting (0, 0) and (3, f(3)). Again the concavity of f shows that f(3)/3 > f(4)/4.

2.2, #10. For f(x) = x^{3} + 5

\frac{f(1 + h) − f(1)}{h} = \frac{((1 + h)^{3} + 5) − (1 + 5)}{h} = \frac{3h + 3h^{2} + h^{3}}{h} → 3

as h → 0.

2.2, #12. For g(t) = 3t^{2} + 5t

\frac{g(−1 + h) − g(−1)}{h} = \frac{(3(−1 + h)^{2} + 5(−1 + h)) − (3 − 5)}{h} = \frac{−h + 3h^{2}}{h} → −1

as h → 0.

Comments: In problems #6, 7 the quantities in question are related to slopes of lines. In #6 the lines are secant lines through two given points on the graph; in #7 the lines are lines connecting the origin to given points on the graph.