AP Calculus AB - Ultimate Guide

Limits and Continuity

Limits

  • Definition: The value that a function approaches as the variable gets nearer to a particular value.
  • Focus is on what's happening around the point, not at the point.

Ways to Find Limits:

  • Graphically: Observe the value the function approaches on the graph.
    • If the graph approaches two different values for the same number, the limit does not exist.
  • Numerically: Estimate from a table of values.
  • Algebraically: Use algebraic manipulation.
    • Factor the numerator and denominator and cancel any removable discontinuities.
    • Useful when the denominator equals 0.
    • Example: \frac{(x+3)(x+2)}{(x+3)(x-3)} The (x+3) term can be removed, indicating a removable discontinuity.
  • Limit of trigonometric functions:
    • \lim_{x \to 0} \frac{\sin(x)}{x} = 1
    • \lim_{x \to 0} \frac{\cos(x) - 1}{x} = 0
    • \lim_{x \to 0} \frac{\sin(ax)}{x} = a
    • \lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)} = \frac{a}{b}

Squeeze Theorem

  • Conditions:
    • For all x in an interval containing a, g(x) \leq f(x) \leq h(x).
    • g and h have the same limit as x approaches a.
    • \lim{x \to a} g(x) = L and \lim{x \to a} h(x) = L, then \lim_{x \to a} f(x) = L

Continuity

Types of Discontinuities:

  • Jump Discontinuity: The curve "breaks" and starts at another point.
    • Limits from the left and right exist but do not match.
  • Essential/Infinite Discontinuity: Curve has a vertical asymptote.
  • Removable Discontinuity: A hole in an otherwise continuous curve.
    • Can be "removed" by filling the hole, often by factoring.

Continuity Conditions

  • For f(x) to be continuous at x = c:
    • f(c) exists.
    • \lim_{x \to c} f(x) exists.
    • \lim_{x \to c} f(x) = f(c)
  • A function is continuous on an interval if it is continuous at every point in that interval.

Removing Discontinuities

  • Redefine the function to exclude the problem point from the domain.
  • Often done by factoring out a common root between the numerator and denominator.

Limits and Asymptotes

  • Vertical Asymptote: A line that a function cannot cross because the function is undefined there.
  • Horizontal Asymptote: Describes the end behavior of a function. Can be crossed.

Horizontal Asymptote Rules

  • If the highest power of x in a rational expression is in the numerator, the limit as x approaches infinity is infinity (no horizontal asymptote).
  • If the highest power of x is in the denominator, the limit as x approaches infinity is zero, and the horizontal asymptote is the line y = 0.
  • If the highest power is the same in the numerator and denominator, the limit is the coefficient of the highest term in the numerator divided by the coefficient of the highest term in the denominator.

Intermediate Value Theorem (IVT)

  • If a function f(x) is continuous on the interval [a, b] and C is any number between f(a) and f(b), then there is at least one number in the interval [a, b] such that f(x) = C.

Differentiation: Definition and Fundamental Properties

Rates of Change

  • Average Rate of Change: Found using the difference quotient over an interval of time.
    • Formula: \frac{y2 - y1}{x2 - x1}.
  • Instantaneous Rate of Change: Rate of change at a specific point in time.
    • Difference quotient with a limit as h \to 0.

Slopes of Lines & Definition of Derivative

  • For a non-linear line, we can use the secant line to approximate the slope.
  • The closer the points are, the more accurate the slope will be.
  • The tangent line- that touches the curve at exactly one point
  • We get this line by using the Instantaneous Rate of Change (remember: the difference quotient but with a limit as h → 0)

Derivative Notation

FunctionFirst DerivativeSecond Derivative
f(x)f'(x)f''(x)
g(x)g'(x)g''(x)
yy' or \frac{dy}{dx}y''
  • The derivative is the rate of change at a specific point, found using a limit as x gets infinitesimally small.

Derivative Rules

  • Constant Rule: If f(x) = k (where k is a constant), then f'(x) = 0.
    • Example: If f(x) = 10, then f'(x) = 0.
  • Constant Multiple Rule: If you have a constant multiplied by a function, you can "pull the constant out."
    • (k * f(x))' = k * f'(x).
  • The Power Rule: If f(x) = x^n, then f'(x) = nx^{n-1}.
    • Example: x^4 becomes 4x^3, and 2x^2 becomes 4x.
    • "Multiply down and decrease the power."
    • Works for polynomials.
  • The Product Rule: If f(x) = uv, then f'(x) = (u)(\frac{dv}{dx}) + (v)(\frac{du}{dx}).
    • Take the first term and multiply it by the derivative of the second term, then add that to the second term multiplied by the derivative of the first term.
    • "1d2 + 2d1" (first * derivative of second + second * derivative of the first)
  • The Quotient Rule: If f(x) = \frac{u}{v}, then f'(x) = \frac{(v)(\frac{du}{dx}) - (u)(\frac{dv}{dx})}{v^2}.
    • Take the denominator and multiply it by the derivative of the numerator, then subtract the numerator multiplied by the derivative of the denominator, all over the denominator squared.
    • "low d high - high d low/ low^2" (low * derivative of high - high * derivative of low over low^2)
  • Memory Derivatives: Derivatives of common functions.
    • \frac{d}{dx} \sin(x) = \cos(x)
    • \frac{d}{dx} \cos(x) = -\sin(x)
    • \frac{d}{dx} e^x = e^x
    • \frac{d}{dx} \ln(x) = \frac{1}{x}

Differentiation: Composite, Implicit, and Inverse Functions

Chain Rule

  • Used when finding the derivative of a composite function.
  • Take the derivative of the outside function with the inside function considered as the variable, leaving the "inside" function alone. Then multiply this by the derivative of the inside function, with respect to the variable x.
  • If y = f(g(x)), then y' = f'(g(x)) * g'(x) OR If y = y(v) and v = v(x), then \frac{dy}{dx} = (\frac{dy}{dv})*(\frac{dv}{dx})
  • Memory trick: "douter, inner, dinner) → drop the power down to outside the parathesis, leave the inner, multiply by the derivative of the inner

Implicit Differentiation

  • Used when you can't isolate y in terms of x.
  • Solve for the derivative of x with respect to y, in order to get a derivative in terms of both variables.
  • (\frac{dx}{dy}) = (\frac{1}{\frac{dy}{dx}})
  • Allows to separate the variable and pair it to both sides, so that they can be factored.
  • The AP exam loves to do this.
  • If your variable doesn’t match dx, then you need to follow it up with d(variable)/dx
  • Example: Given x^2 + y^2 = 25 at the point (3, 4), differentiate implicitly with respect to x:
    • 2x + 2y(\frac{dy}{dx}) = 0
    • Solve for \frac{dy}{dx}: \frac{dy}{dx} = -\frac{x}{y}
    • At the point (3, 4), \frac{dy}{dx} = -\frac{3}{4}.
    • Tangent line equation: y - 4 = -\frac{3}{4}(x - 3)

Inverse Function Differentiation

  • There is a simple formula in order to find the derivative of an inverse function.
  • Find the derivative at a particular point by taking the reciprocal of the derivative at that point’s corresponding y value.
  • If you want g'(1). So find f'(2), then take the reciprocal - this is the value of g'(1).
  • Original point is (1,2) so our reciprocal will be (2,1)

Inverse Trigonometry

  • Easier to just memorize, but you can find them by following the formulas explained in implicit differentiation and using trigonometry rules.

Hints

  • When two terms are multiplied together, use product rule unless it’s easier to multiply it out
  • If you see a function within another function, you will almost certainly have to use chain rule

Contextual Applications of Differentiation

Interpreting the Derivative

  • The derivative tells us the slope of the line tangent to the graph- which tells us the slope of the line at a particular point This means that they can tell us the change of a unit over time.

Straight Line Motion

  • Position: x(t) (sometimes wrote as s(t)) - Meters
  • Velocity: x'(t) or v(t) - Meters/Second
  • Acceleration: x''(t) or v'(t) or a(t) - Meters/Second^2
  • Deriving position gives velocity; deriving velocity gives acceleration.
  • Particles speed up when the signs of velocity and acceleration match (both positive or both negative).
  • Example: If a particle moves along a straight line with velocity function v(t) = 3t^2 - 4t + 2. Find the acceleration of the particle at time t=2?
    • a(t) = \frac{d}{dt} v(t) = 6t - 4
    • a(2) = 6(2) - 4 = 8

Non-Motion Changes

  • The derivative can also tell us the change of something other than motion.
  • Example: Volume of water in a pool is V(t) = 8t^2 -32t +4, where V is the volume in gallons and t is the time in hours.
    • Rate of change of volume: \frac{dV}{dt} = 16t - 32
    • At t=2, the volume isn’t changing (equation equals 0)
  • Example: temperature of a cup of coffee is given by the function x(t) = 70 + 50e^{-0.1t}, where t is the time in minutes since the coffee was poured.
    • d/dt of x(t) = -5e^{-0.1t}
    • Evaluating this derivative at t=5 minutes, we get: x’(5) = -5e^{-0.1(5)} ≈ -2.27

Related Rates

  • Problems where the change of one thing is related to another.
  • Example: A pool of water is expanding at 16\pi square inches per second; find the rate of the radius expanding when the radius is 4 inches.
    • Area: A = \pi r^2
    • Related rates: \frac{dA}{dt} = 2 \pi r (\frac{dr}{dt}) (implicit differentiation).
    • 16\pi = 2\pi(4)\frac{dr}{dt}
    • \frac{dr}{dt} = 2 (radius is changing at a rate of 2 inches per second).
  • Example: a spherical balloon is being inflated at a rate of 10 cubic inches per second. How fast is the radius of the balloon increasing when the radius is 4 inches?
    • We know that the volume of a sphere is given by the formula V = (\frac{4}{3})\pi r^3.
    • \frac{dV}{dt} = 4\pi r^2 (\frac{dr}{dt})
    • 10 = 4\pi(4^2)(\frac{dr}{dt})
    • \frac{dr}{dt} = \frac{10}{(16\pi)}

Steps for Solving Related Rates Problems

  1. Read the problem carefully and identify all given information.
  2. Draw a diagram if possible.
  3. Determine what needs to be found and assign a variable to it.
  4. Write an equation that relates the variables involved.
  5. Differentiate both sides of the equation with respect to time.
  6. Substitute in the given values and solve for the unknown rate.
  7. Remember to always include units in your final answer and to check that your answer makes sense in the context of the problem.

Linearization

  • Differentials are very small quantities that correspond to a change in a number. We use Δx to denote a differential.
  • Approximate the value of a function.
  • f(x + Δx) ≈ f(x) + f'(x)Δx
  • "The value of a function (at x plus a little bit) is equal to the value of the function (at x) plus the product of the derivative of the function (at x) plus a little bit”
  • Example: Approximate (3.98)^4
    • Let f(x) = x^4, x = 4, and Δx = -0.02.
    • f'(x) = 4x^3
    • (x + Δx)^4 = x^4 + 4x^3Δx
    • Plugging in x = 4 and Δx = -0.02 gives 250.88.

L’Hospital’s Rule

  • If a limit gives you \frac{0}{0} or \frac{\infty}{\infty}, then it is called “indeterminate,” and you can use L’Hospital’s Rule to interpret it!
  • L’Hospital’s Rule says that we can take the derivative of the numerator and denominator and try again
  • Example : lim_{x \to \infty} \frac{5x^3 -4x^2 +1}{7x^3 +2x - 6}
    • This equals \frac{\infty}{\infty} so we can take the derivative of the top and bottom
    • \frac{15x^2 -8x}{21x^2 +2}
    • \frac{30x -8}{42x}
    • \frac{30}{42}
    • Answer: \frac{5}{7}

Analytical Applications of Differentiation

The Mean Value Theorem (MVT)

  • The MVT links the average rate of change and the instantaneous rate of change.
  • There must be some point in the interval where the slope of the tangent line equals the slope of the secant line (that connects the endpoints).
  • Rolle’s Theorem is a special case of the MVT

Extreme Value Theorem

  • If a function is continuous on a closed interval, then it must have both a maximum and a minimum value on that interval.
    • Absolute (global) → the function extremum over the entire domain
    • Relative (local) → the function extremum over a particular interval

Intervals of Increase and Decrease

  • Using the first derivative,
  • For all values where f'(x) > 0, the function is increasing.
  • For all values where f'(x) < 0, the function is decreasing.
  • To find the interval of increase and decrease:
    1. Take the derivative of the function
    2. Set it equal to zero to find your critical numbers
    3. Plug in a number above the critical point and below the critical point to find the sign of f'(x)
  • Example: function f(x) = x^3 - 6x^2 + 9x + 2, is increasing or decreasing.
    • derivative f'(x) = 3x^2 - 12x + 9
    • Setting f'(x) = 0 get : 3x^2 - 12x + 9 = 0
    • Solve for x we get: x = 1 or x = 3
    • When x < 1, f'(x) is positive, so the function is increasing.
    • When 1 < x < 3, f'(x) is negative, so the function is decreasing.
    • When x > 3, f'(x) is positive, so the function is increasing.
    • function f(x) is increasing on the intervals (-∞, 1) and (3, ∞) and decreasing on the interval (1, 3).

Relative Extrema

  • The first derivative also tells us relative maxima and minima of a function
  • When the first derivative shifts from positive to negative there will be a relative maximum
  • When the first derivative shifts from negative to positive there will be a relative maximum

Candidate’s Test & Absolute Extrema

  • To find the absolute (global) extrema you have to consider the endpoints and critical numbers
  • Make a table of these values
  • Then plug back into your ORIGINAL FUNCTION

Function Concavity

  • The first derivative tells us if the function is increasing or decreasing, The second derivative tells us if this is happening at an increasing or decreasing rate
  • For all the values that f''(x) > 0, the function is concave up.
  • For all the values that f''(x) < 0, the function is concave down.
  • To find the intervals of concavity and inflection point, the steps are:
    1. Take the second derivative of the function
    2. Set it equal to zero to find your points of inflection
    3. Plug in a number above the critical point and below the critical point to find the sign of f''(x)
  • Example: Consider f(x) = x^3 - 6x^2 + 9x + 2
    • f'(x) = 3x^2 - 12x + 9
    • f''(x) = 6x - 12
    • To find the critical points, we set f''(x) = 0: 6x - 12 = 0
    • x = 2
    • When x < 2, f''(x) < 0, so the function is concave down.
    • When x > 2, f''(x) > 0, so the function is concave up.

Integration and Accumulation of Change

The Integral & Area Under A Curve

  • The integral ∫ also called the antiderivative.
  • The derivative shows us the change/unit So the antiderivative shows us the total change
    The first type is called a definite integral and shows us the area of the region under the function and the x-axis. It gives us the accumulation/total change!
  • To find the definite integral needs us to get the area under the function

Riemann & Trapezoidal Sums

  • We can split this area up into rectangles, the more rectangles we have, the better our estimate is!
  • We can take a Riemann Sum from the left, or from the right!
  • For left-handed sums, we use the endpoints (number) on the left
  • For right-handed sums, we use the endpoints (number) on the right!
  • The formulas are the same for any rectangle, base * height!
  • Take the width of your rectangle and multiply it by the height of the rectangle!Do this for each rectangle you have and add them all together
  • To get these rectangles even more accurate, we can use a midpoint sum, We still use the formula for a rectangle, but we use the value for the height in between!
  • A shape that would more closely fit the shape of the curve is a trapezoid Therefore, we can use trapezoidal sums!
  • Formula for a trapezoid is (1/2)(b1 + b2)(h)

Tabular Riemann Sums

  • The majority of the time when you have to use a Riemann Sum, the AP gives it to you in tabular format
  • Trapezoids: (1/2)(1.5+2)(1) + (1/2)(2+6)(2) + (1/2)(6+11)(2) + (1/2)(11+15)(3)
  • Left Sum: (1)(1.5) + (2)(2) + (2)(6) + (3)(11)
  • Right Sum: (1)(2) + (2)(6) + (2)(11) + (3)(15)

Fundamental Theorem of Calculus & Antiderivatives

  • The same applies for the antiderivative, It’s the opposite of what you do to take a derivative EXCEPT… We can only use the power rule!
  • The power rule for a derivative tells us to multiply down and decrease the power, then the opposite of that would be to divide and increase the power!
  • The +C is very important!
  • If you take the derivative of any number without an x, you get zero Therefore if we’re doing the reverse process, we don’t know what this number could be, therefore we add on a C for the constant of integration and called the constant of integration.
  • integral of 2x is really 2x^2/2 but that simplifies to x^2
  • If the integral is not in power rule format, we must algebraically manipulate it so that we can use the power rule

Advanced Integration

  • Sometimes, getting an integral into power rule format is nearly impossible, in those cases there are other techniques we can do!
  • If your integral contains trigonometry, the best thing to do is just memorize the derivative of trig functions, and the integral will be the opposite.
  • d/dx sinx = cosx
  • \int cosx = sinx
  • Or U-substitution
    1. Chose a term to be your “u”
    2. Take the derivative of this value to get du/dx
    3. Substitute in your u value for the term and your du/dx value for dx
    4. Take the integral
  • Ex. \int(x - 4)^{10}
    1. Let u = x-4
    2. \frac{du}{dx} = 1
    3. dx = \frac{du}{1}
    4. \int(u)^{10} du
    5. \frac{u^{11}}{11} + C
    6. \frac{(x-4)^{11}}{11} + C

Differential Equations

Introduction & Slope Fields

  • Oftentimes, variables are not constant, so we have to represent their change using a derivative. Ex. the change in y is dy/dx
  • A differential equation models the change in one variable with respect to another
  • Slope fields show us what the slopes look like at points on a graph
  • \frac{dy}{dx} = x
  • plug in x/y values into your differential equation and draw that as your slope.
  • The slope at x = -1 would be -1
  • to sketch a solution curve given a slope field, Just “flow” with the slopes

Differential Equations

  • If you’re given a differential equation where the derivative of a function is equal to some other function, you have to solve for the original!
  • You can do this by taking the integral (antiderivative) of both sides!
  • SIPPY problems
    • S: separate (dy and dx on separate sides)
    • I: integrate (remove the derivative)
    • P: Plus C (add your c value to your integral)
    • P: Plug in your initial condition
    • Y: Y equals (solve to find what y is)
  • Example: If \frac{dy}{dx} = \frac{4x}{y} and y(0) = 5 we need to solve for y
    • Start by separating \rightarrow ydy = 4xdx
    • Then integrate \rightarrow \int ydy = \int 4xdx \rightarrow \frac{y^2}{2} = 2x^2 + C
    • Plug in \rightarrow \frac{(5)^2}{2} = 2(0)^2 + C \rightarrow C = \frac{25}{2}
    • Now set y equals \rightarrow y = 2x^2 + \frac{25}{2}

Applications of Integration

Average Value of Functions

  • Remember that to calculate the average we add everything up and then divide!
  • For example, if we had the interval 0 to 40, we can take the integral of our function and divide it by our interval!
    So it would be \frac{1}{40} * \int f(x)

Position, Velocity, and Acceleration

  • Similar to how we can go from position → velocity using the derivative, we can do the reverse using the integral!
  • velocity = \int a(t)
  • Displacement \int v(t)
  • Position \int |v(t)|

Area Between Two Curves

  • The integral gives us the area below a function Therefore, we can subtract the area of one function and another to get the area between the two!
  • Integrate the top function & subtract the bottom function!
  • We need to take the integral from where the functions start (normally zero) to where they intersect
  • \int 5x-x^2 from 0 to 4 - \int x from 0 to 4

Volume by Cross Sectional Area

  • We get a 2D shape from the area under a curve, if we rotate this shape → we get a 3D object
  • To find the area we just integrate the volume formula!
  • V = \int (length x width) dx
  • That would give us the volume for a rectangle (because their area is just l * w)
  • We can almost always use the disc method to find our volume
  • We know that the area of a circle is \pi r^2 so using our integral we would have V = \int \pi r^2
  • \int \pi R^2 - \int \pi r^2