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Home Limits, Analytic Geometry, and Approximations Limits and Curve Sketching Examples Example 5
 
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Example 5

f(x) = x3/5.

Then

f'(x) = 3/5x-2/5, f"(x) = -6/25x-7/5.

At the point x = 0, f(x) = 0 and f'(x) does not exist. We first plot a few points, compute the necessary limits, and make a table.

f(x)

f'(x)

f"(x)

Comments

limx→-∞

- ∞

0

horizontal

x = - 1

-1

3/5

6/25

increasing, ∪

limx→0-

0

∞

vertical

x = 0

0

undef.

limx→0+

0

∞

vertical

x = 1

1

3/5

-6/25

increasing, ∩

limx→∞

∞

0

horizontal

Figure 5.3.3 is a sketch of the curve.

05_limits_g_approx-186.gif

Figure 5.3.3

The behavior as x approaches -∞, ∞, and zero are described by the limits we have computed. As x approaches either - ∞ or ∞, f(x) gets large but the slope becomes more nearly horizontal. As x approaches zero the curve becomes nearly vertical, increasing from left to right, so we have a vertical tangent line at x = 0.
Home Limits, Analytic Geometry, and Approximations Limits and Curve Sketching Examples Example 5

Last Update: 2006-11-15