AP CALCULUS BC TEST BANK EXAM
QUESTIONS AND SOLUTIONS GRADED A+
◍ Average Rate of Change. Answer: Slope of secant line between two
points, use to estimate instantanous rate of change at a point.
◍ Instantenous Rate of Change. Answer: Slope of tangent line at a
point, value of derivative at a point
◍ Formal definition of derivative. Answer:
◍ Alternate definition of derivative. Answer: limit as x approaches a
of [f(x)-f(a)]/(x-a)
◍ When f '(x) is positive, f(x) is. Answer: increasing
◍ When f '(x) is negative, f(x) is. Answer: decreasing
◍ When f '(x) changes from negative to positive, f(x) has a. Answer:
relative minimum
◍ When f '(x) changes from positive to negative, f(x) has a. Answer:
relative maximum
,◍ When f '(x) is increasing, f(x) is. Answer: concave up
◍ When f '(x) is decreasing, f(x) is. Answer: concave down
◍ When f '(x) changes from increasing to decreasing or decreasing to
increasing, f(x) has a. Answer: point of inflection
◍ When is a function not differentiable. Answer: corner, cusp,
vertical tangent, discontinuity
◍ Product Rule. Answer: uv' + vu'
◍ Quotient Rule. Answer: (uv'-vu')/v²
◍ Chain Rule. Answer: f '(g(x)) g'(x)
◍ y = x cos(x), state rule used to find derivative. Answer: product rule
◍ y = ln(x)/x², state rule used to find derivative. Answer: quotient rule
◍ y = cos²(3x). Answer: chain rule
◍ Particle is moving to the right/up. Answer: velocity is positive
, ◍ Particle is moving to the left/down. Answer: velocity is negative
◍ absolute value of velocity. Answer: speed
◍ y = sin(x), y' =. Answer: y' = cos(x)
◍ y = cos(x), y' =. Answer: y' = -sin(x)
◍ y = tan(x), y' =. Answer: y' = sec²(x)
◍ y = csc(x), y' =. Answer: y' = -csc(x)cot(x)
◍ y = sec(x), y' =. Answer: y' = sec(x)tan(x)
◍ y = cot(x), y' =. Answer: y' = -csc²(x)
◍ y = sin⁻¹(x), y' =. Answer: y' = 1/√(1 - x²)
◍ y = cos⁻¹(x), y' =. Answer: y' = -1/√(1 - x²)
◍ y = tan⁻¹(x), y' =. Answer: y' = 1/(1 + x²)
◍ y = cot⁻¹(x), y' =. Answer: y' = -1/(1 + x²)
QUESTIONS AND SOLUTIONS GRADED A+
◍ Average Rate of Change. Answer: Slope of secant line between two
points, use to estimate instantanous rate of change at a point.
◍ Instantenous Rate of Change. Answer: Slope of tangent line at a
point, value of derivative at a point
◍ Formal definition of derivative. Answer:
◍ Alternate definition of derivative. Answer: limit as x approaches a
of [f(x)-f(a)]/(x-a)
◍ When f '(x) is positive, f(x) is. Answer: increasing
◍ When f '(x) is negative, f(x) is. Answer: decreasing
◍ When f '(x) changes from negative to positive, f(x) has a. Answer:
relative minimum
◍ When f '(x) changes from positive to negative, f(x) has a. Answer:
relative maximum
,◍ When f '(x) is increasing, f(x) is. Answer: concave up
◍ When f '(x) is decreasing, f(x) is. Answer: concave down
◍ When f '(x) changes from increasing to decreasing or decreasing to
increasing, f(x) has a. Answer: point of inflection
◍ When is a function not differentiable. Answer: corner, cusp,
vertical tangent, discontinuity
◍ Product Rule. Answer: uv' + vu'
◍ Quotient Rule. Answer: (uv'-vu')/v²
◍ Chain Rule. Answer: f '(g(x)) g'(x)
◍ y = x cos(x), state rule used to find derivative. Answer: product rule
◍ y = ln(x)/x², state rule used to find derivative. Answer: quotient rule
◍ y = cos²(3x). Answer: chain rule
◍ Particle is moving to the right/up. Answer: velocity is positive
, ◍ Particle is moving to the left/down. Answer: velocity is negative
◍ absolute value of velocity. Answer: speed
◍ y = sin(x), y' =. Answer: y' = cos(x)
◍ y = cos(x), y' =. Answer: y' = -sin(x)
◍ y = tan(x), y' =. Answer: y' = sec²(x)
◍ y = csc(x), y' =. Answer: y' = -csc(x)cot(x)
◍ y = sec(x), y' =. Answer: y' = sec(x)tan(x)
◍ y = cot(x), y' =. Answer: y' = -csc²(x)
◍ y = sin⁻¹(x), y' =. Answer: y' = 1/√(1 - x²)
◍ y = cos⁻¹(x), y' =. Answer: y' = -1/√(1 - x²)
◍ y = tan⁻¹(x), y' =. Answer: y' = 1/(1 + x²)
◍ y = cot⁻¹(x), y' =. Answer: y' = -1/(1 + x²)
