AP Calculus AB and BC Course and Exam Description This is the core document for the course. It clearly lays out the course content and describes the exam and AP Program in general. 9 AP Calculus Review Applications #3 Name Calculator DatePd. Grass clippings are placed in a bin, where they decompose. For 0ddt, the amount of grass clippings remaining in the bin is modeled by At) t, where At is measured in pounds and t is measured in days. Class Navigation « Pre-Calculus Review Course for the AP Calculus AB/BC Exam #9 AP Microeconomics Course 1: Basic Economic Concepts and Supply & Demand Analysis #7 ».

Test your readiness for the AP Calculus exam with this quiz!

Answer 1

A: If the graph of a function has a sharp point, the function is not differentiable at that point. This is because the slopes directly to the left and right of the point do not approach the same value.

An absolute value function has a sharp point at its vertex. The graph off(x)=|x + 4| is a horizontal translation (to the left 4 units) of the standard absolute value function, y =|x|, which has vertex (0, 0). Thus, the vertex of fis (-4, 0), which means the function is not differentiable at x =-4. Choice (A) is correct.

Answer 2

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D: The only discontinuities that are removable are holes and holes with a point above or below—this function has neither. The function has 2 jump discontinuities (gaps), at x=-2 and x= 0, but neither of these is removable.

Answer 3

D: Examine the values in the table: f(x) increases as x gets larger, which indicates that f′(x), the slope of the function, is positive. This means f′(x)> 0, so eliminate (A) and (B).

To choose between (C) and (D), take a closer look at the slopes. The slope between the first pair of points is 3, and the slope between the second pair of points is also 3, so f′(x) is constant. This means f′′(x), which is the derivative of f′(x), must be 0. Choice (D) is correct

Answer 4

C: To use a local linear approximation, you need to find the equation of the tangent line. You’ve been given all the information you need in the question stem; you just need to piece it all together. The point on the function is given by f(-1)= 5, which translates to the point (-1, 5). The slope of the tangent line at x=-1 is given by f'(-1)= 2, so the slope is 2. Now the point-slope form of the tangent line is:

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y-5 = 2(x-(-1))

y = 5+2(x+1)

Substituting x=-0.9 into the equation of the tangent line yields:

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5+2(-0.9+1) = 5+2(0.1) = 5.2

That’s (C).

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