AP Calculus BC

\(f(x) = \frac{1}{x}\)

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 \(x \geq 0\).

A function can be continuous from one side.

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. \(q(x) \neq 0\).

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.

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

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.

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.

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

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 \(f(c) = k\).

\(\lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} = f'(x)\)

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.

\(f(x) = k\)

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}

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

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

\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}

\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}

\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}

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

\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}

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

\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}

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}