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Chapter 3 | Derivatives
Example 3.6 Revisiting the Derivative
For f ( x ) =3 x 2 −4 x +1, find f ′(2) by using Equation 3.6 .
Solution Using this equation, we can substitute two values of the function into the equation, and we should get the same value as in Example 3.5 . f ′(2) = lim h →0 f (2+ h )− f (2) h Apply the definition. = lim h →0 (3(2+ h ) 2 −4(2+ h )+1)−5 h Substitute f (2) =5and f (2+ h ) =3(2+ h ) 2 −4(2+ h )+1. = lim h →0 3 h 2 +8 h h Simplify the numerator. = lim h →0 h (3 h +8) h Factor the numerator. = lim h →0 (3 h +8) Cancel the common factor. =8 Evaluate the limit. The results are the same whether we use Equation 3.5 or Equation 3.6 .
For f ( x ) = x 2 +3 x +2, find f ′(1).
3.3
Velocities and Rates of Change Now that we can evaluate a derivative, we can use it in velocity applications. Recall that if s ( t ) is the position of an object moving along a coordinate axis, the average velocity of the object over a time interval [ a , t ] if t > a or [ t , a ] if t < a is given by the difference quotient (3.7) v ave = s ( t )− s ( a ) t − a . As the values of t approach a , the values of v ave approach the value we call the instantaneous velocity at a . That is, instantaneous velocity at a , denoted v ( a ), is given by (3.8) v ( a ) = s ′( a ) = lim t → a s ( t )− s ( a ) t − a . To better understand the relationship between average velocity and instantaneous velocity, see Figure 3.8 . In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time t = a whose position at time t is given by the function s ( t ). The slope of the secant line (shown in green) is the average velocity of the object over the time interval [ a , t ].
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