Calculus AB Main Points

Unit 1 Main Points

Two-Sided Limits

  • Existence of Two-Sided Limits: For the limit extlim<em>xoaf(x)=Lext{lim}<em>{x o a} f(x) = L to exist, the left-hand limit and the right-hand limit must be equal:   - extlim</em>xoa−f(x)=Lext{lim}</em>{x o a^-} f(x) = L and extlimxoa+f(x)=Lext{lim}_{x o a^+} f(x) = L

Value and Limit of a Function

  • The value of a function at a point aa is independent of the limit as xx approaches aa.
      - For example, if f(a)=Lf(a) = L, it does not imply that extlimxoaf(x)=Lext{lim}_{x o a} f(x) = L holds.
  • A function can be undefined at x=ax = a while extlimxoaf(x)ext{lim}_{x o a} f(x) can still exist.

Limit Laws

  • For a constant cc and limits extlim<em>xoaf(x)ext{lim}<em>{x o a} f(x) and extlim</em>xoag(x)ext{lim}</em>{x o a} g(x), the following limit laws apply:
      - Sum Rule: extlim<em>xoa[f(x)+g(x)]=extlim</em>xoaf(x)+extlim<em>xoag(x)ext{lim}<em>{x o a} [f(x) + g(x)] = ext{lim}</em>{x o a} f(x) + ext{lim}<em>{x o a} g(x)   - Constant Multiple Rule: extlim</em>xoa[cf(x)]=cextlim<em>xoaf(x)ext{lim}</em>{x o a} [c f(x)] = c ext{lim}<em>{x o a} f(x)   - Difference Rule: extlim</em>xoa[f(x)−g(x)]=extlim<em>xoaf(x)−extlim</em>xoag(x)ext{lim}</em>{x o a} [f(x) - g(x)] = ext{lim}<em>{x o a} f(x) - ext{lim}</em>{x o a} g(x)
      - Product Rule: extlim<em>xoa[f(x)g(x)]=extlim</em>xoaf(x)imesextlim<em>xoag(x)ext{lim}<em>{x o a} [f(x) g(x)] = ext{lim}</em>{x o a} f(x) imes ext{lim}<em>{x o a} g(x)   - Quotient Rule: extlim</em>xoaf(x)g(x)=extlim<em>xoaf(x)extlim</em>xoag(x)ext{lim}</em>{x o a} \frac{f(x)}{g(x)} = \frac{ ext{lim}<em>{x o a} f(x)}{ ext{lim}</em>{x o a} g(x)} if extlim<em>xoag(x)≠0ext{lim}<em>{x o a} g(x) \neq 0   - Power Rule: extlim</em>xoa[f(x)]n=(extlimxoaf(x))next{lim}</em>{x o a} [f(x)]^n = ( ext{lim}_{x o a} f(x))^n

Methods for Finding Limits

Direct Substitution

  • Direct substitution can be applied when the function is continuous at the point approaching.
  • Usable for the following types of functions:
      1. Polynomials
      2. Rational functions
      3. Trigonometric functions

Example Scenarios

  1. Existence of Limit:
       - For extlimxo2ext√x−2x−2ext{lim}_{x o 2} \frac{ ext{√}x - 2}{x - 2}, secures that its limit does exist.
  2. Non-Existence of Limit:
       - For extlimxo2ext√x−2x−2ext{lim}_{x o 2} \frac{ ext{√}x - 2}{x-2}, depending on conditions, the limit might be DNE (Does Not Exist).

Indeterminate Forms

Rewriting Techniques

  • Four algebra techniques allow rewriting functions in indeterminate forms:
  1. Factoring and Canceling: Find common factors to simplify.
  2. Common Denominators: Change expressions to a common denominator.
  3. Multiplying By the Conjugate: Apply conjugate pairs to eliminate square roots.
  4. Multiply Out and Simplify: Distributing and combining like terms appropriately.

End Behavior and Asymptotes

Horizontal Asymptotes
  • To determine horizontal asymptotes:
      - If extlimxoextinfinityf(x)=cext{lim}_{x o ext{infinity}} f(x) = c, then y=cy = c is a horizontal asymptote.
Vertical Asymptotes
  • Existence indicated when substituting a point x=ax = a leads the denominator to zero:
      - If extlimxoaf(x)=ext±ext∞ext{lim}_{x o a} f(x) = ext{±} ext{∞}, then there is a vertical asymptote at x=ax=a.

Types of Discontinuities

Forms of Discontinuity

  • Removable Discontinuity (Hole): Exists at x=cx=c if making f(c)f(c) continuous “fills in the hole” (extending ff).
  • Jump Discontinuity: Exists if left and right-hand limits at x=ax=a both exist but differ.
  • Infinite Discontinuity: If approaching aa drives f(a)f(a) to infinity (vertical asymptote).

Continuity at a Point

  • A function f(x)f(x) is continuous at x=ax=a iff:
      1. f(a)f(a) is defined
      2. extlim<em>xoaf(x)ext{lim}<em>{x o a} f(x) exists   3. extlim</em>xoaf(x)=f(a)ext{lim}</em>{x o a} f(x) = f(a)

Unit 2 Main Points

Definition of the Derivative

  • For function f(x)f(x) with respect to variable xx, the derivative f′(x)f'(x) is defined as:
    f′(x)=extlimho0f(x+h)−f(x)hf'(x) = ext{lim}_{h o 0} \frac{f(x + h) - f(x)}{h} provided that limit exists.

Slope Interpretations

  • Average Rate of Change: Represents the slope of the secant line.
  • Instantaneous Rate of Change: Represents the slope of the tangent line at a specific point.

Differentiability and Continuity

  • If a function is differentiable at a point, it must be continuous there, but continuity alone doesn't guarantee differentiability.
      - Example functions highlight continuous yet non-differentiable (e.g., f(x)=∣x∣f(x) = |x| at x=0x=0).

Rules of Derivatives

Rules of Differentiation

  1. Constant Function: ddx(c)=0\frac{d}{dx}(c) = 0
  2. Power Rule: ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}
  3. Constant Multiple Rule: ddx[cf(x)]=cddx[f(x)]\frac{d}{dx}[c f(x)] = c \frac{d}{dx}[f(x)]
  4. Sum Rule: ddx[f(x)+g(x)]=f′(x)+g′(x)\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)
  5. Difference Rule: ddx[f(x)−g(x)]=f′(x)−g′(x)\frac{d}{dx}[f(x) - g(x)] = f'(x) - g'(x)
  6. Product Rule: ddx[f(x)g(x)]=f′(x)g(x)+f(x)g′(x)\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)
  7. Quotient Rule: ddx[f(x)g(x)]=f′(x)g(x)−f(x)g′(x)g(x)2\frac{d}{dx}[\frac{f(x)}{g(x)}] = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}

Derivatives of Trigonometric Functions

\ - ddx(extsinx)=extcosx\frac{d}{dx}( ext{sin } x) = ext{cos } x
\ - ddx(extcosx)=−extsinx\frac{d}{dx}( ext{cos } x) = - ext{sin } x
\ - ddx(exttanx)=extsec2x\frac{d}{dx}( ext{tan } x) = ext{sec}^2 x
\ - ddx(extsecx)=extsecxanx\frac{d}{dx}( ext{sec } x) = ext{sec } x an x
\ - ddx(extcscx)=−extcscxextcotx\frac{d}{dx}( ext{csc } x) = - ext{csc} x ext{cot } x
\ - ddx(extcotx)=−extcsc2x\frac{d}{dx}( ext{cot } x) = - ext{csc}^2 x

Higher Order Derivatives

  • Second Derivative: f′′(x)f''(x) or d2ydx2\frac{d^2y}{dx^2}
  • Third Derivative: f′′′(x)f'''(x) or d3ydx3\frac{d^3y}{dx^3}
  • nth Derivative: f(n)(x)f^{(n)}(x) or dnydxn\frac{d^ny}{dx^n}

The Chain Rule

  • The chain rule is expressed in multiple forms:
      1. dydx=f′(u)dudx\frac{dy}{dx} = f'(u) \frac{du}{dx}
      2. If h(x)=f(g(x))h(x) = f(g(x)), then h′(x)=f′(g(x))g′(x)h'(x) = f'(g(x)) g'(x).

Particle Motion

  • Displacement: Calculated as s(b)−s(a)s(b) - s(a)
  • Average Velocity: s(b)−s(a)b−a\frac{s(b) - s(a)}{b - a}
  • Instantaneous Velocity: v(t)=s′(t)v(t) = s'(t)
  • Acceleration: a(t)=v′(t)=s′′(t)a(t) = v'(t) = s''(t)

Motion Dynamics

  • Condition defining movement direction:
      - Moving left if v(t)<0v(t) < 0   - Moving right if v(t)>0v(t) > 0
  • Condition for speeding up and slowing down is dependent on the sign of velocity and acceleration.

Implicit Differentiation

  1. Differentiate both sides of an equation w.r.t. xx.
  2. Collect terms containing yy on one side.
  3. Factor out y′y' and solve for it.

Unit 6 AP Calculus AB Main Points

Area Between Two Curves

  • Area A between curves y=f(x)y=f(x) and y=g(x)y=g(x) from x=ax=a to x=bx=b:
    A=extArea=extlimoextvolume=extintegralfunction=extfunctionboundsA = ext{Area} = ext{lim} o ext{volume} = ext{integral function} = ext{function bounds}
  • The formula for area is expressed as:
    A=extArea=ext∫fromatobof(f(x)−g(x))extdxA = ext{Area} = ext{∫ from a to b of} (f(x) - g(x)) ext{d}x for continuous and defined between aa and bb.

Average Value of a Function

  • The average value of a function ff on interval [a,b][a,b] is given by:
    fextavg=1b−aext∫fromatoboff(x)extdxf_{ ext{avg}} = \frac{1}{b - a} ext{∫ from a to b of } f(x) ext{d}x

Unit 7 AP Calculus AB Main Points

Volumes of Solids Using Disk and Washer Methods.

  • Volume of Solid (disk method - horizontal axis of revolution):
    V=extπext∫fromatobof[f(x)]2extdxV = ext{π} ext{∫ from a to b of } [f(x)]^2 ext{d}x
  • Washer Method (area between functions):
    V=extπext∫fromatobof[R(x)2−r(x)2]extdxV = ext{π} ext{∫ from a to b of } [R(x)^2 - r(x)^2] ext{d}x

Solids of Known Cross-sections

  • Volume for cross-sections perpendicular to the x-axis:
    V=ext∫fromatobofA(x)extdxV = ext{∫ from a to b of } A(x) ext{d}x
  • Volume for cross-sections perpendicular to the y-axis:
    V=ext∫fromctodofA(y)extdyV = ext{∫ from c to d of } A(y) ext{d}y

Differential Equations and Population Growth

  • For equations of the form dydt=ky\frac{dy}{dt} = ky, the solution is exponential:
    y(t)=y0ekty(t) = y_0 e^{kt}, where y0y_0 is the initial value.