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Example 1

For the curve

y = x3,

find the tangent lines at the points (0,0), (1,1), and (-½, -⅛) (Figure 2.2.2).

The slope is given by

f'(x) = 3x2.

02_differentiation-46.gif

Figure 2.2.2

At x = 0,

f'(0) = 3 · 02 = 0.

The tangent line has the equation

y = 0(x - 0) + 0, or y = 0.

At x = 1,

f'(1) = 3,

whence the tangent line is

y = 3(x - 1) + 1, or y = 3x - 2.

At x = -½,

f'(-½) = 3 -(-½)2 = ,

so the tangent line is

y = (x - (-½)) + (-⅛), or y = x +


Last Update: 2006-11-15