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2014 AP CALCULUS AB FRQs. Average rate of change = lbs/day. On the fifteenth day the amount of grass clippings is decreasing at the rate of 0.1635 lbs /day. Average amount of clippings: Solve days. Tangent line at is Solve days. y = 4. f (x).

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2014 AP CALCULUS AB FRQs

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2014 AP CALCULUS AB FRQs


Average rate of change = lbs/day


On the fifteenth day the amount of grass clippings is decreasing at the rate of 0.1635 lbs/day


Average amount of clippings:

Solvedays


Tangent line at is

Solve days


y= 4

f(x)

Intersections of the two graphs: and


y= 4

f(x)

Length of the rectangle:

y – f(x)


y= 4

f(x)


y= 4

f(x)


y= 4

f(x)


, therefore if g(x) increases . This happens on the intervals and (−3, 2)

If g(x) i decreases. This happens on and (0, 4)

g(x)is increasing and concave down on and (0, 2)


The slope of the graph of f at is −2, so


Average acceleration of train A:

S meters/minute2


Since is differentiable, it is also continuous so the Intermediate Value Theorem applies to for . Therefore, there must be at least a value such that meters/minute


meters


Train B

meters

meters

meters

z = distance between the two trains

y

Train A

x

At minutes,


f has a local minimum when its derivative, f’, switches from negative to positive. This occurs at x = 1.


f is twice differentiable so its derivative, f’, is both continuous and differentiable. Therefore the Mean Value Theorem can be used on f’ .

There must be a value for such that

Since


The problem can also be done using Rolle’s Theorem: f is twice differentiable so its derivative, f’, is both continuous and differentiable.

And since Rolle’s Theorem can be used on f’.

Therefore, there must be a value for such that


Equation of the tangent line:


Using (0, 1):


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