AP Calculus AB Exam

• Intermediate Value Theorem


  If f(1)=-4 and f(6)=9, then there must be a x-value between 1 and 6 where f crosses the x-axis.
• Average Rate of Change


  Slope of secant line between two points, use to estimate instantanous rate of change at a point.
• Instantenous Rate of Change


  Slope of tangent line at a point, value of derivative at a point
• Formal definition of derivative


  limit as h approaches 0 of [f(a+h)-f(a)]/h
• Alternate definition of derivative


  limit as x approaches a of [f(x)-f(a)]/(x-a)
• When f '(x) is positive, f(x) is

  increasing
• When f '(x) is negative, f(x) is

  decreasing
• When f '(x) changes from negative to positive, f(x) has a

  relative minimum
• When f '(x) changes fro positive to negative, f(x) has a

  relative maximum
• When f '(x) is increasing, f(x) is

  concave up
• When f '(x) is decreasing, f(x) is


  concave down
• When f '(x) changes from increasing to decreasing or decreasing to increasing, f(x) has a

  point of inflection
• When is a function not differentiable

  corner, cusp, vertical tangent, discontinuity
• Product Rule

  uv' + vu
• Quotient Rule


  (uv'-vu')/v²
• Chain Rule

  f '(g(x)) g'(x)
• y = x cos(x), state rule used to find derivative

  product rule
• y = ln(x)/x², state rule used to find derivative

  quotient rule
• y = cos²(3x)


  chain rule
• Particle is moving to the right/up

  velocity is positive
• Particle is moving to the left/down

  velocity is negative
• absolute value of velocity


  speed
• y = sin(x), y' =


  y' = cos(x)
• y = cos(x), y' =

  y' = -sin(x)
• y = tan(x), y' =

  y' = sec²(x)
• y = csc(x), y' =

  y' = -csc(x)cot(x)
• y = sec(x), y' =

  y' = sec(x)tan(x)
• y = cot(x), y' =

  y' = -csc²(x)
• y = sin⁻¹(x), y' =

  y' = 1/√(1 - x²)
• y = cos⁻¹(x), y' =

  y' = -1/√(1 - x²)
• y = tan⁻¹(x), y' =

  y' = 1/(1 + x²)
• y = cot⁻¹(x), y' =

  y' = -1/(1 + x²)
• y = e^x, y' =


  y' = e^x
• y = a^x, y' =


  y' = a^x ln(a)
• y = ln(x), y' =

  y' = 1/x
• y = log (base a) x, y' =


  y' = 1/(x lna)
• To find absolute maximum on closed interval [a, b], you must consider...

  critical points and endpoints
• mean value theorem

  if f(x) is continuous and differentiable, slope of tangent line equals slope of secant line at least once in the interval (a, b)f '(c) = [f(b) - f(a)]/(b - a)
• If f '(x) = 0 and f"(x) > 0,

  f(x) has a relative minimum
• If f '(x) = 0 and f"(x) < 0,

  f(x) has a relative maximum