Free-Response on Application of Derivatives

1. 2011 # 4 Free-Response on Application of Derivatives Nina Luksanapol

4. (a) Find g(-3). Find g’(x) and evaluate g’(-3). g’(x)= 2 + f(x) g’(-3)= 2 + f(-3) = 2+ 0 = 2

5. (b) Determine the x-coordinate of the point at which g has an absolute maximum on the interval -4 x 3.Justify your answer. To find an abolute maximum you have to find where g’(x) = 0. g’(x) = 2 + f(x) = 0 So f(x) must equal -2 f(x) = -2 at x= 2.5 Therefore, g(x) has an absolute maximum at x= 2.5 X= 2.5 f(x) = -2

6. (c) Find all values of x on the interval -4 < x < 3 for which the graph of g has a point of inflection. Give a reason for your answer. To find the point of inflection you must set g’’(x) = 0. g’’(x) = f’(x) = 0 There is technically no point on the graph where f’(x) = 0. Since at x = 0, the point is undifferentiable. However, since you are finding the point of inflection of g(x), you are finding the point where the sign of f’(x) changes, where at x= 0, the sign changes from positive to negative. So, the point of inflection for g(x) is at x = 0

7. (d) Find the average rate of change of f on the interval -4 x 3. There is no point c, -4 < c < 3, for which f’(c) is equal to that average rate of change. Explain why this statement does not contradict the Mean Value Theorem. Average Rate of Change: _1__ 3-(-4) f’(x) dx = [ f(3) – f(-4) ] = [ -3 – (-1) ] = 1 7 1 7 -2 7 This statement does not contradict the Mean Value Theorem because x = -4 and x = 3 is not differentiable. It is also not differentiable at x = 0. And there is no f’(c)= f(3) – f(-4) 3 – (-4)

8. Citations! apcentral.collegeboard.com http://home.roadrunner.com/~askmrcalculus/help.html

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