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Chapter 4 | Applications of Derivatives
of c are guaranteed, first calculate the derivative of f . The derivative f ′( x ) = 1 (2 x ) .
The slope of the line
connecting (0, f (0)) and (9, f (9)) is given by f (9)− f (0) 9−0 =
9− 0 9−0 =
3 9 =
1 3 .
We want to find c such that f ′( c ) = 1 3 .
That is, we want to find c such that
1 2 c = 1 3 .
Solving this equation for c , we obtain c = 9 4 .
At this point, the slope of the tangent line equals the slope of the
line joining the endpoints.
Figure 4.27 The slope of the tangent line at c =9/4 is the same as the slope of the line segment connecting (0, 0) and (9, 3).
One application that helps illustrate the Mean Value Theorem involves velocity. For example, suppose we drive a car for 1 h down a straight road with an average velocity of 45 mph. Let s ( t ) and v ( t ) denote the position and velocity of the car, respectively, for 0≤ t ≤1 h. Assuming that the position function s ( t ) is differentiable, we can apply the Mean Value Theorem to conclude that, at some time c ∈ (0, 1), the speed of the car was exactly v ( c ) = s ′( c ) = s (1)− s (0) 1−0 =45mph. Example 4.16 Mean Value Theorem and Velocity
If a rock is dropped from a height of 100 ft, its position t seconds after it is dropped until it hits the ground is
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