Chapter 3 | Derivatives
219
f (3+ h )− f (3) h
m tan = lim h →0
Apply the definition.
(3+ h ) 2 −9 h
f (3+ h ) = (3+ h ) 2 and f (3) =9.
= lim
Substitute
h →0
9+6 h + h 2 −9 h
= lim
Expand and simplify to evaluate the limit.
h →0
h (6+ h ) h
(6+ h ) =6
= lim
= lim
h →0
h →0
We obtained the same value for the slope of the tangent line by using the other definition, demonstrating that the formulas can be interchanged.
Example 3.3 Finding the Equation of a Tangent Line
Find the equation of the line tangent to the graph of f ( x ) =1/ x at x =2.
Solution We can use Equation 3.3 , but as we have seen, the results are the same if we use Equation 3.4 . m tan = lim x →2 f ( x )− f (2) x −2 Apply the definition. = lim x →2 1 x − 1 2 x −2 Substitute f ( x ) = 1 x and f (2) = 1 2 .
1 x − 1 2 x −2 ·
Multiply numerator and denominator by2 x to simplify fractions.
2 x 2 x
= lim
x →2
(2− x ) ( x −2)(2 x )
= lim
Simplify.
x →2
Simplify using 2− x
−1 2 x
= lim
x ≠2.
x −2 =−1, for
x →2
= − 1 4
Evaluate the limit.
We now know that the slope of the tangent line is − 1 4 .
To find the equation of the tangent line, we also need a
point on the line. We know that f (2) = 1 2 . we can use the point-slope equation of a line to find the equation of the tangent line. Thus the tangent line has the equation y = − 1 4 x +1. The graphs of f ( x ) = 1 x and y = − 1 4 x +1 are shown in Figure 3.7 . Since the tangent line passes through the point (2, 1 2 )
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