AP Calculus BC

Table of Contents

1 Continuity

Continuity of \(f\) over a domain \(D\) means that for all \(x\) in \(D\) there exists an \(f(x)\).

Continuity at point c means that there is a limit at \(c\), a defined \(f(c)\), and \(\lim\limits_{x \to c} f(x) = f(c)\).

1.1 Examples

\(f(x) = \frac{1}{x}\)
This function is continuous at all points but \(x = 0\).
\(f(x) = \sqrt[n]{x}\) where \(n\) is any positive integer
This function is continuous
Where \(n\) is odd
Over all points.
Where \(n\) is even
Over all points where \(x \geq 0\).

1.2 One sided limits

A function can be continuous from one side.

1.2.1 Examples

\(f(x) = \sqrt{n^{2} - x}\)
Semicircle function. Continuous from the right at \(-n\) and from the left at \(n\).

1.3 Properties of continuity

Compositions
For a continuous function \(f\) and a continuous function \(g\), \(f \circ g\) and \(g \circ f\) are both continuous. If however, \(f\) is continuous but \(g\) is discontinuous, \(f \circ g\) is discontinuous, but \(g \circ f\) can be continuous if \(f\)'s range does not include any points of discontinuity in \(g\).
Polynomials
\(f(x) = x\) is continuous, so any multiplicative or additive operation on x produces a continuous function.
Rationals
Rationals , \(\frac{p(x)}{q(x)}\), are continuous on all points where the denominator is not equal to zero, i.e. \(q(x) \neq 0\).

1.4 Removable discontinuities

If a function has a hole in it, for example, \(f(x) = \frac{x^{2} - 1}{x - 1}\) and it can be removed by limiting, for example,

\begin{align} \frac{x^{2} - 1}{x - 1} &= \frac{ \left(x + 1\right)\left(x - 1\right) }{ x - 1 } \\ \lim\limits_{x \to -1} \frac{x^{2} - 1}{x - 1} &= x + 1 \end{align}

The discontinuity at \(x = -1\) has been removed via a limit.

1.5 Nonremovable discontinuities

If a function attemps to reach a point, but there is no limit for the undefined value there, it is called a nonremovable discontinuity.

1.5.1 Examples

\(f(x) = \frac{1}{x}\)
At \(x = 0\), \(\frac{1}{x}\) is undefined. Since both sides approach different values, there is no limit, and so, no way to remove the discontinuity.
\(f(x) = \sin{\frac{1}{x}}\)
If it is defined that \(f(x) = \left(g \circ h\right)(x)\), \(g(x) = \sin{x}\) and \(h(x) = \frac{1}{x}\). Since \(h(x)\) has a nonremovable discontinuity, by the rules laid out earlier, \(f(x)\) has a nonremovable discontinuity at \(x = 0\).

2 Continuity (cont.)

Continuity rules:

  1. All polynomials are continuous for all values of \(x\).
  2. Fractions in the form of \(h(x) = \frac{f(x)}{g(x)}\) are discontinuous wherever \(g(x) = 0\). It makes no difference if \(g(x)\) cancels out. When it does, it just creates a removable discontinuity.
  3. Radicals in the form of \(g(x) = \sqrt[\text{odd root}]{f(x)}\) are continuous everywhere.
  4. Radicals in the form of \(g(x) = \sqrt[\text{even root}]{f(x)}\) are discontinuous when \(f(x) < 0\).
  5. Trig functions in the form of \(f(x) = \sin{ax}\) or \(f(x) = \cos{ax}\) are continuous everywhere.
  6. Exponentials in the form of \(f(x) = a^{bx}\) are continuous everywhere.

2.1 Examples

Find discontinuities:

\(f(x) = -3x^{2} - 5x + 1\)
Continuous, as it's a polynomial.
\(f(x) = \frac{x^{2} - 1}{x - 1}\)
A removable discontinuity at \(x = 1\) due to the numerator and denominator evaluating to \(0\) and the fraction can be simplified using a limit.
\(f(x) = \sqrt[3]{x^{2} + 2x - 4}\)
Cube root (continuous) composition of a polynomial (continuous). Continuous due to the composition rules laid out earlier.
\(f(x) = \sqrt{x^{2} - x - 6}\)
The inner polynomial can be factored to \((x + 2)(x - 3)\) which, due to a positive largest coefficient, \(1\), is negative from \((-2, 3)\) and as such the square root is discontinuous upon that range.
\(f(x) = \frac{\cos{\left(x + 2\right)}}{x - 2}\)
Cosine is continuous. It is divided by \(x - 2\) which is \(0\) at \(x = 2\). This means the function is discontinuous there.
\(f(x) = e^{-x} - 1\)
Continuous as all exponential functions are continuous.

2.2 Wtih piecewise functions

While all individual functions in the piecewise definition can be continuous, this does not mean the whole piecewise function is continuous. The rule change points need to be checked for discontinuities.

\begin{equation} f(x) = { \begin{cases} x^{2} - 3 & x \leq 1 \\ 1 - x & x > 1 \end{cases} } \end{equation}

The rule change lies at \(x = 1\). Since both of the rules are polynomials, they are continuous. Since at \(x = 1\), \(x^{2} - 3 = -2\) and \(1 - x = 0\), the \(f(x)\) is not continuous.

\begin{equation} f(x) = { \begin{cases} x^{2} + 3x - 2 & x \leq -2 \\ -x^{2} & x > -2 \end{cases} } \end{equation}

The rule change here is \(x = -2\). Again, both rules are polynomials. At \(x = -2\), \(x^{2} + 3x - 2 = -4\) and \(-x^{2} = -4\), meaning the function is continuous.

\begin{equation} f(x) = { \begin{cases} 3^{-x} & x \leq -1 \\ \frac{1}{x + 1} & x > -1 \end{cases} } \end{equation}

This rule change is at \(x = -1\). The rational has a discontinuity at \(x = -1\) that goes to infinity. Therefore, the function is not continuous.

\begin{equation} f(x) = { \begin{cases} \frac{x - 4}{x^{2} - 16} & x \neq 4 \\ \frac{1}{3x - 4} & x = 4 \end{cases} } \end{equation}

The rule change is at \(x = 4\) due to \(\frac{x - 4}{x^{2} - 16}\) having a removable discontinuity there. By removing the discontinuity with a limit and checking the second rule's value for equality, continuity can be verified.

\begin{align} \frac{x - 4}{x^{2} - 16} &= \frac{ x - 4 }{ \left(x + 4\right)\left(x - 4\right) } \\ \lim\limits_{x \to 4} \frac{x - 4}{x^{2} - 16} &= \frac{1}{x + 4} \end{align}

Now, by inputting \(x = 4\), \(\frac{1}{x + 4} = \frac{1}{8}\) and \(\frac{1}{3x - 4} = \frac{1}{8}\), so \(f\) is continuous.

2.2.1 Practice with \(k\)

Some piecewise functions may require you to find \(k\) to make the function continuous. To solve for \(k\), try to make both rules equal to eachother at the rule change. Remove any removable discontinuities not in the rule's domain.

\begin{equation} f(x) = { \begin{cases} 3x + 2 & x \leq 1 \\ 2k - x & x > 1 \end{cases} } \end{equation}

Setting \(x = 1\) and making the two expressions equal results in \(5 = 2k - 1\).

\begin{align} 5 &= 2k - 1 \\ (+1) && (+1) \\ 6 &= 2k \\ (\div 2) && (\div 2) \\ 3 &= k \end{align}

Inserting \(k = 3\), the rules become \(3x - 2\) and \(6 - x\).

\begin{equation} f(x) = { \begin{cases} kx^{2} & x \leq 2 \\ kx - 6 & x > 2 \end{cases} } \end{equation}

Again, the rule change is at \(x = 2\). The expressions become \(4k\) and \(2k - 6\).

\begin{align} 4k &= 2k - 6 \\ (-2k) && (-2k) \\ 2k &= -6 \\ (\div 2) && (\div 2) \\ k &= -3 \end{align}

Inserting the new \(k = -3\), the rules become \(-3x^{2}\) and \(-3x - 6\).

3 Theorems

Important theorems for Calculus.

3.1 Theorem Rules

Basic theorems follow a format of "If \(x\), then \(y\)".

3.1.1 Converse

Theorems have a converse if they follow "If and only if". "If and only if \(x\), then \(y\)" becomes "If and only if \(y\), then \(x\)."

3.1.2 Negation

Theorems have a negative if they follow "If and only if". "If and only if \(x\), then \(y\)" becomes "If and only if not \(x\), then not \(y\)."

3.1.3 Contrapositive

All theorems can have a contrapositive. The contropositive is a combination of the converse and the negation. "If \(x\), then \(y\)" becomes "If not \(y\), then not \(x\)".

Contrapositives change logic of \(x\) and \(y\). Optionals become required, "And" becomes "Or", and vice versa.

3.2 Intermediate Value Theorem

The Intermediate Value Theorem states that if \(f\) is continuous on a closed interval \([a, b]\), a \(c\) exists for any \(k\) where \(k\) is between \(f(a)\) and \(f(b)\) and \(f(c) = k\).

If the \(f\) is not continuous on the closed interval, then there can be a \(k\) where there is no \(c\) such that \(f(c) = k\).

3.2.1 Conditional statement

If
there is at least one number \(c\) in \([a, b]\) such that \(f(c) = k\)
Then
\(f\) is continuous on the closed interval \([a, b]\), \(f(a) \neq f(b)\) or \(k\) is any number between \(f(a)\) and \(f(b)\).

3.2.2 Usage: Finding x-intercepts of functions

Given a function \(f\), if \(f(a)\) is positive and \(f(b)\) is negative, there is some \(c\) where \(f(c) = 0\).

\(f(x) = x^{3} + x - 1\)
Here, \(f(-1) = -1\) and \(f(1) = 1\), so given an interval \([-1, 1]\) there must be a \(c\) such that \(f(c) = 0\). Solving for \(c\), set the equation to \(0\).
\(f(x) = x^{3} + 5x - 3\)
Here, \(f(0) = -3\) and \(f(1) = 3\). By the IVT, there is a \(0 < c < 1\) such that \(f(c) = 0\).
\(g(t) = 2\cos{t} - 3t\)
Here, \(g(0) = 2\) and \(g\left(\frac{\pi}{2}\right) = -\frac{3\pi}{2}\), so there is at least one \(c\) where \(0 < c < \frac{\pi}{2}\) and \(g(c) = 0\).
\(h(\theta) = 1 + \theta - 3\tan{\theta}\)
Here, \(h(0) = 1\) and \(h\left(\frac{\pi}{4}\right) \approx -\frac{5}{4}\). This means that there is a \(c\) in \(\left[0, \frac{\pi}{4}\right]\) where \(h(c) = 0\).

3.2.3 Usage: Finding a value inside a set interval

Given a function \(f\), a closed interval \([a, b]\), and a \(k\), prove that there is a \(c\) such that \(f(c) = k\) and \(a < c < b\).

\(f(x) = x^{2} + x - 1\), \([0, 5]\), \(f(c) = 11\)

Substituting \(c\) and setting equal to \(11\):

\begin{align} & c^{2} + c - 1 &= 11 & \\ & c^{2} + c - 12 &= 0 & \\ & c &= \frac{-1 \pm \sqrt{1^{2} - 4(1)(-12)}}{2} & \\ & c &= \frac{-1 \pm \sqrt{49}}{2} & \\ & c &= \frac{-1 \pm 7}{2} & \\ c &= \frac{-8}{2} & c &= \frac{6}{2} \\ c &= -4 & c &= 3 \end{align}

\(-4\) is not in the domain, so the solution is \(c = 3\).

\(f(x) = x^{2} - 6x + 8\), \([0, 3]\), \(f(c) = 0\)

Again, substitute \(c\) and set equal to \(0\).

\begin{align} & c^{2} - 6c + 8 &= 0 & \\ & c &= \frac{6 \pm \sqrt{(-6)^{2} - 4(1)(8)}}{2} & \\ & c &= \frac{6 \pm \sqrt{36 - 32}}{2} & \\ & c &= \frac{6 \pm \sqrt{4}}{2} & \\ & c &= \frac{6 \pm 2}{2} & \\ c &= \frac{4}{2} & c &= \frac{8}{2} \\ c &= 2 & c &= 4 \end{align}

Since \(4\) is not in \([0, 3]\), the solution is \(c = 2\).

\(f(x) = x^{3} - x^{2} + x - 2\), \([0, 3]\), \(f(c) = 4\)

Again, substitute \(c\) and set equal to \(f(c)\).

\begin{align} c^{3} - c^{2} + c - 2 &= 4 \\ c^{3} - c^{2} + c - 6 &= 0 \\ (c - 2)\left(c^{2} + x + 3\right) &= 0 \\ c &= 2 \end{align}

The left over quadratic is not worth solving for solutions as due to the cubic not being factorable further said solutions would not be real. The result is \(c = 2\).

\(f(x) = \frac{x^{2} + x}{x - 1}\), \(\left[\frac{5}{2}, 4\right]\), \(f(c) = 6\)
The rational has a nonremovable discontinuity at \(x = 1\), but \(x = 1\) is not included in our domain, so the function is continuous over the domain. Guessing integers inside of the domain leads to \(c = 3\). Author note: I've forgotten how to solve rationals.

4 Derivatives

For a function \(f(x)\), the derivative, \(f'(x)\), is a new function that outputs the slope of \(f(x)\) at \(x\).

4.1 Definitions

\(\lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} = f'(x)\)
This definition represents a point approaching \((x, f(x))\) along \(f(x)\) and a secant line being taken between the two points.
\(\lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(x)\)
This is the same as the above definition except that instead of a difference limiting to \(0\), \(x\) limits.

4.2 Taking example limits

\(f(x) = k\)

Set up the limit and solve.

\begin{align} f'(x) &= \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ f'(x) &= \lim\limits_{h \to 0} \frac{k - k}{h} \\ f'(x) &= \lim\limits_{h \to 0} 0 \\ f'(x) &= 0 \end{align}
\(f(x) = x^{n} \left[n \in \mathbb{Z}\right]\)

Set up the limit, solve using binomial distribution.

\begin{align} f'(x) &= \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ f'(x) &= \lim\limits_{h \to 0} \frac{(x + h)^{n} - x^{n}}{h} \\ f'(x) &= \lim\limits_{h \to 0} \frac{\left((x + h) - x\right)\left( x^{n - 1} + x^{n - 2}(x + h) + x^{n - 3}(x + h)^{2} + \ldots + x^{2}(x + h)^{n - 3} + x(x + h)^{n - 2} + (x + h)^{n - 1}\right) }{h} \\ f'(x) &= \lim\limits_{h \to 0} \frac{h\left( x^{n - 1} + x^{n - 2}(x + h) + x^{n - 3}(x + h)^{2} + \ldots + x^{2}(x + h)^{n - 3} + x(x + h)^{n - 2} + (x + h)^{n - 1}\right) }{h} \\ f'(x) &= \lim\limits_{h \to 0} x^{n - 1} + x^{n - 2}(x + h) + x^{n - 3}(x + h)^{2} + \ldots + x^{2}(x + h)^{n - 3} + x(x + h)^{n - 2} + (x + h)^{n - 1} \\ f'(x) &= x^{n - 1} + x^{n - 1} + x^{n - 1} + \ldots + x^{n - 1} + x^{n - 1} + x^{n - 1} \\ f'(x) &= nx^{n - 1} \end{align}

4.3 Theorem

If \(f(x)\) is differentiable, then \(f(x)\) is continuous.

If \(f(x)\) is not continuous, then \(f(x)\) is not differentiable.

5 Derivatives (cont.)

More derivative rules:

5.1 Constant Multiple Rule

\begin{align} g(x) &= k f(x) \\ g'(x) &= \lim\limits_{h \to 0} \frac{g(x + h) - g(x)}{h} \\ g'(x) &= \lim\limits_{h \to 0} \frac{k f(x + h) - k f(x)}{h} \\ g'(x) &= \lim\limits_{h \to 0} \left[k \left(\frac{f(x + h) - f(x)}{h}\right)\right] \\ g'(x) &= k \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ g'(x) &= k f'(x) \end{align}

5.2 Sum and Difference Rule

\begin{align} t(x) &= u(x) + v(x) \\ t'(x) &= \lim\limits_{h \to 0} \frac{t(x + h) - t(x)}{h} \\ t'(x) &= \lim\limits_{h \to 0} \frac{u(x + h) - u(x) + v(x + h) - v(x)}{h} \\ t'(x) &= \lim\limits_{h \to 0} \left(\frac{u(x + h) - u(x)}{h} + \frac{v(x + h) - v(x)}{h}\right) \\ t'(x) &= \lim\limits_{h \to 0} \left(\frac{u(x + h) - u(x)}{h}\right) + \lim\limits_{h \to 0} \left(\frac{v(x + h) - v(x)}{h}\right) \\ t'(x) &= u'(x) + v'(x) \end{align}

5.3 Product Rule

\begin{align} t(x) &= u(x) \cdot v(x) \\ t'(x) &= \lim\limits_{h \to 0} \frac{t(x + h) - t(x)}{h} \\ t'(x) &= \lim\limits_{h \to 0} \frac{u(x + h)v(x + h) - u(x)v(x)}{h} \\ t'(x) &= \lim\limits_{h \to 0} \frac{u(x + h)v(x + h) + u(x)v(x + h) - u(x)v(x + h) - u(x)v(x)}{h} \\ t'(x) &= \lim\limits_{h \to 0} \frac{v(x + h)\left[u(x + h) - u(x)\right] + u(x)\left[v(x + h) - v(x)\right]}{h} \\ t'(x) &= \left(\lim\limits_{h \to 0} v(x + h)\right)\left(\lim\limits_{h \to 0} \frac{u(x + h) - u(x)}{h}\right) + \left(\lim\limits_{h \to 0} u(x + h)\right)\left(\lim\limits_{h \to 0} \frac{v(x + h) - v(x)}{h}\right) \\ t'(x) &= v(x)u'(x) + u(x)v'(x) \end{align}

6 Expressing Derivatives

Newton and Leibniz came up with different ways to express derivatives. Legrange's notation is currently taught as Newton's notation.

If \(y = f(x)\), on the left is Newton (Legrange's notation) and on the right is Leibniz:

\begin{align} f'(x) &= \frac{dy}{dx} \\ f''(x) &= \frac{d^{2}y}{dx^{2}} \\ f'''(x) &= \frac{d^{3}y}{dx^{3}} \\ f^{\left(4\right)} &= \frac{d^{4}y}{dx^{4}} \\ f^{\left(5\right)} &= \frac{d^{5}y}{dx^{5}} \\ \vdots & \vdots \\ f^{\left(n\right)} &= \frac{d^{n}y}{dx^{n}} \end{align}

6.1 Derivatives not expressed by \(x\)

Sometimes, the change in \(y\) over \(x\) is not wanted, but some other dependent is.

6.1.1 The Chain Rule

If \(z\) depends on \(y\) and \(y\) depends on \(x\), the chain rule can be expressed in Leibniz notation, and if \(z = f(y)\) and \(y = g(x)\), then it can be explained in Legrange notation.

\begin{equation} \frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx} = f'(y)g'(x) = f'(g(x))g'(x) \end{equation}

7 More Complex Derivatives

7.1 \(f(x) = \ln{x}\)

\begin{align} f'(x) &= \lim\limits_{h \to 0} \frac{\ln{\left(x + h\right)} - \ln{x}}{h} \\ f'(x) &= \lim\limits_{h \to 0} \frac{1}{h} \left(\ln{\left(x + h\right)} - ln{x}\right) \\ f'(x) &= \lim\limits_{h \to 0} \frac{1}{h} \ln{\left(\frac{x + h}{x}\right)} \\ f'(x) &= \lim\limits_{h \to 0} \frac{1}{h} \ln{\left(1 + \frac{h}{x}\right)} \\ f'(x) &= \lim\limits_{h \to 0} \ln{\left(\left(1 + \frac{h}{x}\right)^{\frac{1}{h}}\right)} \\ [ u &= \frac{h}{x} ] \\ f'(x) &= \lim\limits_{u \to 0} \ln{\left(\left(1 + u\right)^{\frac{1}{ux}}\right)} \\ f'(x) &= \lim\limits_{u \to 0} \frac{1}{x} \ln{\left(\left(1 + u\right)^{\frac{1}{u}}\right)} \\ f'(x) &= \frac{1}{x} \ln{e} \\ f'(x) &= \frac{1}{x} \end{align}

7.2 Inverses

The inverse of a function is defined such that \(x = f^{-1}\left(f(x)\right)\).

7.2.1 Function types valid for inverses

Functions, to have a functional inverse, have to be

Increasing
For all \(x\), \(f(x + d)\) for any positive \(d\) must be greater than \(f(x)\).
Decreasing
For all \(x\), \(f(x + d)\) for any positive \(d\) must be less than \(f(x)\).

7.2.2 Derivative of the Inverse

\begin{align} y &= f^{-1}(x) \\ f(y) &= f\left(f^{-1}(x)\right) \\ f(y) &= x \\ f'(y)\frac{dy}{dx} &= 1 \\ \frac{dy}{dx} &= \frac{1}{f'(y)} \\ \frac{df^{-1}(x)}{dx} &= \frac{1}{f'(y)} \\ f^{-1} {'}(x) &= \frac{1}{f'(y)} \end{align}

7.3 Power Rule for All Real Numbers

\begin{align} f(x) &= x^{n} \\ f'(x) &= \frac{dx^{n}}{dx} \\ f'(x) &= \frac{d\left(e^{\ln{x}}\right)^{n}}{dx} \\ f'(x) &= \frac{de^{n\ln{x}}}{dx} \\ f'(x) &= e^{n\ln{x}} \cdot \frac{n}{x} \\ f'(x) &= \frac{n}{x} e^{\ln{x^{n}}} \\ f'(x) &= \frac{n}{x} x^{n} \\ f'(x) &= \frac{nx^{n}}{x} \\ f'(x) &= nx^{n - 1} \end{align}

8 Optimization

Since first derivatives' zeroes can be used to calculate relative minima and maxima of a function, optimizing a physical problem or a related rate is done that way.

8.1 Cutting Corners for a Box

Given a box that is \(24 \si{in} \times 24 \si{in}\), find the \(x\) that makes the largest volume possible if the corners are cut out in squares of \(x \times x\).

\begin{align} V &= l \cdot w \cdot h \\ V &= \left(24 - 2x\right)\left(24 - 2x\right)x \\ V &= 4x^{3} - 96x^{2} + 576x \\ \frac{dV}{dx} &= 12x^{2} - 192x + 576 \\ \frac{dV}{dx} &= 12\left(x^{2} - 16x + 48\right) \\ \frac{dV}{dx} &= 12\left(x - 12\right)\left(x - 4\right) \\ \frac{dV}{dx} = 0, x &= \cancel{12}, 4 \\ V &= \left(24 - 2\left(4\right)\right) \left(24 - 2\left(4\right)\right)4 \\ V &= 16 \cdot 16 \cdot 4 \\ V &= 1024 \si{in^{3}} \end{align}

9 Antiderivatives

Definition:

If \(F\left(x\right)\) is the antiderivative of \(f\left(x\right)\) then \(F\left(x\right) = \int{f\left(x\right)}{dx}\).

9.1 Rules

9.1.1 Constants

For any constant \(C\):

\begin{align} \frac{d}{dx} C &= 0 \\ \int{0}{dx} &= C \end{align}

9.1.2 Linears

For any constant \(k\):

\begin{align} \frac{d}{dx} kx &= k \\ \int{k}{dx} &= kx + C \end{align}

9.1.3 Constant Multiples

For any function \(f\left(x\right)\) and ant constant \(k\):

\begin{align} \frac{d}{dx} kf\left(x\right) &= kf'\left(x\right) \\ \int{kf'\left(x\right)}{dx} &= kf\left(x\right) + C \end{align}

9.1.4 Powers

\begin{align} \frac{d}{dx} x^{n} &= nx^{n - 1} \\ \int{\left(n + 1\right)x^{n}} &= x^{n + 1} \\ \int{x^{n}} &= \frac{x^{n + 1}}{n + 1} \end{align}

10 Riemann Sums

A Riemann Sum is a way of representing area under a curve, A.K.A a definite integral.

Date: 2019-12-02 Mon

Author: Soren

Created: 2019-12-02 Mon 17:53

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