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Example 7

Sketch the curve 05_limits_g_approx-188.gif for 0 < x < 2π.

f(x) and f'(x) are undefined at x = π because the denominator sin π is zero. The first two derivatives are

05_limits_g_approx-189.gif

Thus f'(x) is always negative, and f"(x) = 0 when x = π/2, 3π/2. Here is the table:

f(x)

f'(x)

f"(x)

Comments

limx→0+

- ∞

vertical

π/4

1

-1/2

+

decreasing, ∪

π/2

0

-1

0

decreasing, inflection

3π/4

-1

-1/2

-

decreasing, ∩

limx→π-

-∞

-∞

vertical

limx→π+

-∞

vertical

5π/4

1

-1/2

+

decreasing, ∪

3π/2

0

-1

0

decreasing, inflection

7π/4

-1

-1/2

-

decreasing, ∩

limx→2π-

-∞

-∞

vertical

Notice that the table from π to 2π is just a repeat of the table from 0 to π. This is because

05_limits_g_approx-190.gif

The curve is sketched in Figure 5.3.5.

05_limits_g_approx-191.gif

Figure 5.3.5


Last Update: 2006-11-14