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AP Calculus BC Exam Review QUESTIONS & ANSWERS 100% CORRECT!!
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Average Rate of Change - ANSWERSlope of secant line between two points, use to estimate instantanous rate of change at a point. Instantenous Rate of Change - ANSWERSlope of tangent line at a point, value of derivative at a point Definition of Derivative - ANSWERlimit as h approaches 0 of [f(a...

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  • January 26, 2025
  • 19
  • 2024/2025
  • Exam (elaborations)
  • Questions & answers

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  • ap calculus bc exam review questions answers 100
  • ap calculus bc exam stuvia 2025
  • average rate of change answerslope of secant li
  • instantenous rate of change answerslope of tange

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AP Calculus BC Exam Review
QUESTIONS & ANSWERS 100%
CORRECT!!
Average Rate of Change - ANSWERSlope of secant line between two points, use to
estimate instantanous rate of change at a point.

Instantenous Rate of Change - ANSWERSlope of tangent line at a point, value of
derivative at a point

Definition of Derivative - ANSWERlimit as h approaches 0 of [f(a+h)-f(a)]/h or limit as
x approaches a of [f(x)-f(a)]/(x-a)

When f '(x) is positive, f(x) is - ANSWERincreasing

When f '(x) is negative, f(x) is - ANSWERdecreasing

When f '(x) changes from negative to positive, f(x) has a - ANSWERrelative minimum

When f '(x) changes fro positive to negative, f(x) has a - ANSWERrelative maximum

When f '(x) is increasing, f(x) is - ANSWERconcave up

When f '(x) is decreasing, f(x) is - ANSWERconcave down

When f '(x) changes from increasing to decreasing or decreasing to increasing, f(x)
has a - ANSWERpoint of inflection

When is a function not differentiable - ANSWERcorner, cusp, vertical tangent,
discontinuity

Product Rule - ANSWERuv' + vu'

Quotient Rule - ANSWER(uv'-vu')/v²

Chain Rule - ANSWERf '(g(x)) g'(x)

Particle is moving to the right/up - ANSWERvelocity is positive

Particle is moving to the left/down - ANSWERvelocity is negative

absolute value of velocity - ANSWERspeed

y = sin(x), y' = - ANSWERy' = cos(x)

y = cos(x), y' = - ANSWERy' = -sin(x)

,y = tan(x), y' = - ANSWERy' = sec²(x)

y = csc(x), y' = - ANSWERy' = -csc(x)cot(x)

y = sec(x), y' = - ANSWERy' = sec(x)tan(x)

y = cot(x), y' = - ANSWERy' = -csc²(x)

y = sin⁻¹(x), y' = - ANSWERy' = 1/√(1 - x²)

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

y = tan⁻¹(x), y' = - ANSWERy' = 1/(1 + x²)

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

y = e^x, y' = - ANSWERy' = e^x

y = a^x, y' = - ANSWERy' = a^x ln(a)

y = ln(x), y' = - ANSWERy' = 1/x

y = log (base a) x, y' = - ANSWERy' = 1/(x lna)

To find absolute maximum on closed interval [a, b], you must consider... -
ANSWERcritical points and endpoints

Linearization - ANSWERuse tangent line to approximate values of the function

left riemann sum - ANSWERuse rectangles with left-endpoints to evaluate integral
(estimate area)

right riemann sum - ANSWERuse rectangles with right-endpoints to evaluate
integrals (estimate area)

Trapezoidal rRle - ANSWERuse trapezoids to evaluate integrals (estimate area)

[(h1 - h2)/2]*base - ANSWERarea of trapezoid

definite integral - ANSWERhas limits a & b, find antiderivative, F(b) - F(a)

indefinite integral - ANSWERno limits, find antiderivative + C, use inital value to find
C

area under a curve - ANSWER∫ f(x) dx integrate over interval a to b

area above x-axis is - ANSWERpositive

area below x-axis is - ANSWERnegative

, average value of f(x) - ANSWER= 1/(b-a) ∫ f(x) dx on interval a to b

To find particular solution to differential equation, dy/dx = x/y - ANSWERseparate
variables, integrate + C, use initial condition to find C, solve for y

To draw a slope field, - ANSWERplug (x,y) coordinates into differential equation,
draw short segments representing slope at each point

slope of horizontal line - ANSWERzero

slope of vertical line - ANSWERundefined

methods of integration - ANSWERsubstitution, parts, partial fractions

∫ u dv = - ANSWERuv - ∫ v du

dP/dt = kP(M - P) - ANSWERlogistic differential equation, M = carrying capacity

P = M / (1 + Ae^(-Mkt)) - ANSWERlogistic growth equation

given rate equation, R(t) and inital condition when
t = a, R(t) = y₁ find final value when t = b - ANSWERy₁ + Δy = y
Δy = ∫ R(t) over interval a to b

given v(t) and initial position t = a, find final position when t = b - ANSWERs₁+ Δs = s
Δs = ∫ v(t) over interval a to b

given v(t) find displacement - ANSWER∫ v(t) over interval a to b

given v(t) find total distance travelled - ANSWER∫ abs[v(t)] over interval a to b

area between two curves - ANSWER∫ f(x) - g(x) over interval a to b, where f(x) is top
function and g(x) is bottom function

volume of solid with base in the plane and given cross-section - ANSWER∫ A(x) dx
over interval a to b, where A(x) is the area of the given cross-section in terms of x

volume of solid of revolution - no washer - ANSWERπ ∫ r² dx over interval a to b,
where r = distance from curve to axis of revolution

volume of solid of revolution - washer - ANSWERπ ∫ R² - r² dx over interval a to b,
where R = distance from outside curve to axis of revolution, r = distance from inside
curve to axis of revolution

length of curve - ANSWER∫ √(1 + (dy/dx)²) dx over interval a to b

indeterminate forms - ANSWER0/0, ∞/∞, ∞*0, ∞ - ∞, 1^∞, 0 ⁰, ∞⁰

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