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2014 AP CALCULUS AB FRQs PowerPoint PPT Presentation

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).

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):