Calculus Study Guide: Rates of Change, Limits, Continuity, and Derivatives

Rates of Change: Average vs. Instantaneous

Understanding rate of change is often more critical in practical applications than knowing isolated values at static points in time.
Average Rate of Change:
- Measures the change in the output variable, on average, for each one-unit increase in the input variable over a specified interval.
- Calculated using algebraic slope methods without calculus.
- Example: Determining how the per-unit cost of production changes on average when output increases from $100$ to $150$ units.
Instantaneous Rate of Change:
- Measures how rapidly the output variable changes at a single, exact input value.
- Requires calculus techniques to evaluate.
- Example: Determining the exact rate of change of per-unit production cost at precisely $100$ units or $150$ units.
Mathematical Notation for Change:
- The Greek letter Delta ( $\Delta$ ) signifies "change in."
- $\Delta x$ represents the change in input ( $\Delta x = b - a$ ).
- $\Delta f$ or $\Delta y$ represents the change in output ( $\Delta f = f(b) - f(a)$ ).

Quantity	Delta Notation
Change in a quantity	$\Delta$
Change in output, $f(x)$	$\Delta f = f(b) - f(a)$
Change in input, $x$	$\Delta x = b - a$

Definition of Average Rate of Change:
- For a function $f$ over the closed interval $[a, b]$ :

$\text{Average rate} = \frac{\text{change in output}}{\text{change in input}} = \frac{\Delta f}{\Delta x} = \frac{f(b) - f(a)}{b - a}$

Secant line showing Delta x and Delta f on a function graph

Secant Lines and Tangent Lines

Secant Line:
- A line connecting any two distinct points $(a, f(a))$ and $(b, f(b))$ on a function's graph.
- The slope of the secant line passing through $(a, f(a))$ and $(b, f(b))$ equals the average rate of change over $[a, b]$ .
Tangency Line (Tangent Line):
- A line that touches a graph at a single point.
- Point of Tangency: The specific coordinate where the tangent line touches the graph.
- As the distance between two points on a curve approaches zero, the secant line approaches the tangent line, and the secant slope approaches the instantaneous rate of change.

$\text{Slope of Tangent Line} = \text{Instantaneous Rate of Change at Point of Tangency}$

Applied Examples of Rates of Change

Example 1: Monthly Profit Function
- Given the profit function $P(q) = -0.05q^2 + 1.2q - 2$ , where $P$ is profit in thousands of dollars and $q$ is production volume in hundreds of units.
- Part A: Average rate of change between $q = 4$ and $q = 10$ :
- Calculate output values: $P(4) = -0.05(4)^2 + 1.2(4) - 2 = -0.05(16) + 4.8 - 2 = 2$ $P(10) = -0.05(10)^2 + 1.2(10) - 2 = -0.05(100) + 12 - 2 = 5$
- Coordinates: $(x_1, y_1) = (4, 2)$ and $(x_2, y_2) = (10, 5)$ .
- Calculate slope: $\frac{\Delta P}{\Delta q} = \frac{P(10) - P(4)}{10 - 4} = \frac{5 - 2}{10 - 4} = \frac{3}{6} = 0.5$
- Interpretation: On average, profit increases by $0.5$ thousand dollars per $100$ units produced over $[4, 10]$ . Thus, between $400$ and $1,000$ units, average profit increases by $\$500$ for each additional $100$ units produced and sold.

Graph of profit function showing secant line between q=4 and q=10

Part B: Average rate of change between $q = 16$ and $q = 20$ :
- Calculate output values: $P(16) = -0.05(16)^2 + 1.2(16) - 2 = -0.05(256) + 19.2 - 2 = 4.4$ $P(20) = -0.05(20)^2 + 1.2(20) - 2 = -0.05(400) + 24 - 2 = 2$
- Coordinates: $(16, 4.4)$ and $(20, 2)$ .
- Calculate slope: $\frac{\Delta P}{\Delta q} = \frac{P(20) - P(16)}{20 - 16} = \frac{2 - 4.4}{4} = \frac{-2.4}{4} = -0.6$
- Interpretation: On average, profit decreases by $0.6$ thousand dollars per $100$ units over $[16, 20]$ . Between $1,600$ and $2,000$ units, profit decreases by $\$600$ for each additional $100$ units produced and sold.

Graph of profit function showing secant line between q=16 and q=20

Example 2: Acme Incorporated Product Revenue
- Weekly revenue model: $R(t) = -0.02t^2 + 2t + 21$ , where $R$ is revenue in millions of dollars and $t$ is time in weeks after product launch.
- Rate of change between week 20 and week 30:
- Calculate values: $R(20) = -0.02(20)^2 + 2(20) + 21 = -8 + 40 + 21 = 53$ $R(30) = -0.02(30)^2 + 2(30) + 21 = -18 + 60 + 21 = 63$
- Average rate of change: $\frac{\Delta R}{\Delta t} = \frac{R(30) - R(20)}{30 - 20} = \frac{63 - 53}{10} = \frac{10}{10} = 1$
- Interpretation: Revenue increased on average by $\$1$ million per week between weeks $20$ and $30$ .
- Numerical Approximation of Instantaneous Rate of Change at $t = 20$ :
- Set fixed point at $(20, R(20)) = (20, 53)$ .
- Evaluate average rate $\frac{\Delta R}{\Delta t} = \frac{R(20 + \Delta t) - R(20)}{(20 + \Delta t) - 20}$ for shrinking positive $\Delta t$ values (approaching from right):

Fixed Point	Second Point	$\Delta t$	Average Rate of Change $\frac{\Delta R}{\Delta t}$
$(20, 53)$	$(21, 54.18)$	$1$	$\frac{54.18 - 53}{21 - 20} = 1.18\,\text{\$million/week}$
$(20, 53)$	$(20.1, 53.1198)$	$0.1$	$\frac{53.1198 - 53}{20.1 - 20} = 1.198\,\text{\$million/week}$
$(20, 53)$	$(20.01, 53.011998)$	$0.01$	$\frac{53.011998 - 53}{20.01 - 20} = 1.1998\,\text{\$million/week}$

* Evaluate average rate for shrinking negative  $\Delta t$  values (approaching from left):

Fixed Point	Other Point	$\Delta t$	Average Rate of Change $\frac{\Delta R}{\Delta t}$
$(20, 53)$	$(19.9, 52.8798)$	$-0.1$	$\frac{52.8798 - 53}{19.9 - 20} = 1.202\,\text{\$million/week}$
$(20, 53)$	$(19.99, 52.987998)$	$-0.01$	$\frac{52.987998 - 53}{19.99 - 20} = 1.2002\,\text{\$million/week}$

* As  $\Delta t \rightarrow 0$ ,  $\frac{\Delta R}{\Delta t} \rightarrow 1.2$ .
* The instantaneous rate of change of revenue at week  $20$  is  $\$1.2$  million per week.

Example 3: Graphical Tangent Line Evaluation
- Given $f(x) = x^2 - 4x + 3$ with a point of tangency at $(1, 0)$ .
- Identify a second point on the tangent line: $(1.5, -1)$ or $(2, -2)$ .
- Calculate tangent slope:

$\text{Slope} = \frac{-1 - 0}{1.5 - 1} = \frac{-1}{0.5} = -2$

The instantaneous rate of change of $f$ at $x = 1$ is $-2$ .

Graph of f(x) = x^2 - 4x + 3 with a tangent line at x=1

Limits: Definitions and One-Sided Behavior

Definition of a Limit:
- Describes the value that a function $f(x)$ approaches as input $x$ approaches a specified value $a$ :

$\lim_{x \rightarrow a} f(x)$

Focuses entirely on the behavior of $f(x)$ near $x = a$ , not the actual value of $f(x)$ at $x = a$ .
- One-Sided Limits:
Left-Hand Limit ( $x \rightarrow a^-$ ): Input $x$ approaches $a$ from values strictly smaller than $a$ :

$\lim_{x \rightarrow a^-} f(x)$

Right-Hand Limit ( $x \rightarrow a^+$ ): Input $x$ approaches $a$ from values strictly greater than $a$ :

$\lim_{x \rightarrow a^+} f(x)$

Graph demonstrating left-hand and right-hand limits approaching L at x=a

Limit Existence Criterion:

$\lim_{x \rightarrow a} f(x) = L \iff \lim_{x \rightarrow a^-} f(x) = L \text{ and } \lim_{x \rightarrow a^+} f(x) = L$

If left-hand and right-hand limits disagree or do not exist, the two-sided limit Does Not Exist (DNE).
Graph Representations:
- Closed Circle (●): Indicates the point is included and the function is defined there.
- Open Circle (○): Indicates a point hole where the function is undefined or redefined.

Limit Calculations and Special Cases

Example 1: Identical Left- and Right-Hand Limits
- Function $f(x) = x + 4$ as $x \rightarrow 0$ :

$x$	$-0.1$	$-0.01$	$-0.001$	$0$	$0.001$	$0.01$	$0.1$
$f(x)$	$3.9$	$3.99$	$3.999$	?	$4.001$	$4.01$	$4.1$

$\lim_{x \rightarrow 0^-} f(x) = 4$ and $\lim_{x \rightarrow 0^+} f(x) = 4$ .
Therefore, $\lim_{x \rightarrow 0} f(x) = 4$
- Example 2: Jump Discontinuity (Non-Existent Limit)
Piecewise function:

$f(x) = \begin{cases} 2x + 1 & \text{if } x < 2 \\ 4 - x & \text{if } x \ge 2 \end{cases}$

Left-hand limit ( $x \rightarrow 2^-$ ):

$x$	$1.9$	$1.99$	$1.999$	$1.9999$
$f(x)$	$4.8$	$4.98$	$4.998$	$4.9998$

$\lim_{x \rightarrow 2^-} f(x) = 5$

Right-hand limit ( $x \rightarrow 2^+$ ):

$x$	$2.1$	$2.01$	$2.001$	$2.0001$
$f(x)$	$1.9$	$1.99$	$1.999$	$1.9999$

$\lim_{x \rightarrow 2^+} f(x) = 2$

Since $\lim_{x \rightarrow 2^-} f(x) = 5 \neq 2 = \lim_{x \rightarrow 2^+} f(x)$ , $\lim_{x \rightarrow 2} f(x) = \text{DNE}$ .

Graph of piecewise function showing jump discontinuity at x=2

Example 3: Infinite Limits and Vertical Asymptotes
- Rational function $f(x) = \frac{5x}{x - 3}$ near $x = 3$ :
- Left-hand values ( $x \rightarrow 3^-$ ):

$x$	$2.9$	$2.99$	$2.999$	$2.9999$	$2.999999999$
$f(x)$	$-145$	$-1495$	$-14995$	$-149995$	$-1.5 \times 10^{10}$

$\lim_{x \rightarrow 3^-} f(x) = -\infty$

Right-hand values ( $x \rightarrow 3^+$ ):

$x$	$3.1$	$3.01$	$3.001$	$3.0001$	$3.000000001$
$f(x)$	$155$	$1505$	$15005$	$150005$	$1.5 \times 10^{10}$

$\lim_{x \rightarrow 3^+} f(x) = \infty$

$\lim_{x \rightarrow 3} f(x) = \text{DNE}$ . A vertical asymptote exists at $x = 3$

Graph of f(x) = 5x / (x - 3) showing vertical asymptote at x=3

Example 4: Limit Exists Where Function is Undefined
- Rational function $f(x) = \frac{x^2 - 9}{x - 3}$ , undefined at $x = 3$ :
- Left-hand limit ( $x \rightarrow 3^-$ ):

$x$	$2.9$	$2.99$	$2.999$	$2.9999$
$f(x)$	$5.9$	$5.99$	$5.999$	$5.9999$

$\lim_{x \rightarrow 3^-} f(x) = 6$

Right-hand limit ( $x \rightarrow 3^+$ ):

$x$	$3.1$	$3.01$	$3.001$	$3.0001$
$f(x)$	$6.1$	$6.01$	$6.001$	$6.0001$

$\lim_{x \rightarrow 3^+} f(x) = 6$

Therefore, $\lim_{x \rightarrow 3} f(x) = 6$ despite $f(3)$ being undefined.

Graph showing open circle at point (3, 6) for f(x) = (x^2 - 9)/(x - 3)

Example 5: Comprehensive Graphical Limit Assessment ( $q(x)$ )
- At $x = -8$ : $\lim_{x \rightarrow -8^-} q(x) = -7$ , $\lim_{x \rightarrow -8^+} q(x) = -7 \implies \lim_{x \rightarrow -8} q(x) = -7$ . Note that $q(-8) = -3$ .
- At $x = -2$ : $\lim_{x \rightarrow -2^-} q(x) = 2$ , $\lim_{x \rightarrow -2^+} q(x) = \infty \implies \lim_{x \rightarrow -2} q(x) = \text{DNE}$ .
- At $x = 6$ : $\lim_{x \rightarrow 6^-} q(x) = 3$ , $\lim_{x \rightarrow 6^+} q(x) = 6 \implies \lim_{x \rightarrow 6} q(x) = \text{DNE}$ .
- At $x = 10$ : $\lim_{x \rightarrow 10^-} q(x) = 2$ , $\lim_{x \rightarrow 10^+} q(x) = 2 \implies \lim_{x \rightarrow 10} q(x) = 2$

Graph of function q(x) used for assessing limits

Mathematical Definition of Continuity

Continuity at a Point:
- A function $f(x)$ is continuous at $x = a$ if and only if all three conditions are satisfied:
1. The limit exists at the point: $\lim_{x \rightarrow a} f(x)$ exists.
2. The function is defined at the point: $f(a)$ exists.
3. The function value equals the limit:

$\lim_{x \rightarrow a} f(x) = f(a)$

Continuity on Intervals and Real Numbers:
- Continuous on an interval: Continuous at every point within that interval.
- Continuous on the real line: Continuous at every real number ( $\mathbb{R}$ ), which includes positive/negative fractions, decimals, and irrational numbers (e.g., $\pi$ , $\sqrt{2}$ ).

Functional Families and Continuous Domains

Functions Continuous Everywhere ( $\mathbb{R}$ ):
- Polynomial Functions:

$f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$

Exponents $n$ are non-negative integers. Example: $f(x) = 0.01x^3 - 2x - 3$ .

Exponential Functions: $f(x) = e^{g(x)}$ where $g(x)$ is polynomial. Example: $g(x) = 5e^{0.02x}$ .
Products of Polynomials and Exponentials: Example: $h(x) = 5e^{0.02x}(0.01x^3 - 2x - 3)$ .
- Functions Continuous Restricted to Their Domain:
Rational Functions: $f(x) = \frac{P(x)}{Q(x)}$ where $P(x), Q(x)$ are polynomials.
- Continuous for all $x$ where $Q(x) \neq 0$ .
- Example: $q(t) = \frac{2t + 3}{4 - t}$ is continuous everywhere except $t = 4$
Radical Functions: $f(x) = \sqrt[n]{g(x)}$ for integer $n > 1$
- Even roots require $g(x) \ge 0$ .
- Example: $r(z) = \sqrt{2z - 6}$ is continuous for $2z - 6 \ge 0 \implies z \ge 3$
Logarithmic Functions: $f(x) = \ln(g(x))$
- Requires argument $g(x) > 0$
- Example: $s(x) = \ln(x - 5)$ is continuous for $x - 5 > 0 \implies x > 5$
- Direct Substitution Property:
If $f(x)$ is continuous at $x = a$ , then limits are evaluated by direct substitution:

$\lim_{x \rightarrow a} f(x) = f(a)$

Examples:
- $\lim_{x \rightarrow 2} (x^2 + 5x - 1) = (2)^2 + 5(2) - 1 = 13$
- $\lim_{t \rightarrow -1} \frac{2t + 3}{4 - t} = \frac{2(-1) + 3}{4 - (-1)} = \frac{1}{5}$
- $\lim_{z \rightarrow 5} e^{0.02z} = e^{0.02(5)} = e^{0.1} \approx 1.105$
- $\lim_{z \rightarrow 4} \sqrt{2z - 6} = \sqrt{2(4) - 6} = \sqrt{2} \approx 1.41$

Classification of Discontinuities

Jump Discontinuity:
- Occurs when $\lim_{x \rightarrow a^-} f(x)$ and $\lim_{x \rightarrow a^+}$ exist but are unequal.
- Example: Rental Car Cost Model
- Rental rate: $\$50$ per day or partial day (no prorating).

$C(t) = 50 \times \lceil t \rceil$

* At  $t = 3$  days:  $\lim_{t \rightarrow 3^-} C(t) = 150$ ,  $\lim_{t \rightarrow 3^+} C(t) = 200$ .
* Limit  $\lim_{t \rightarrow 3} C(t) = \text{DNE}$ . Function  $C(t)$  has a jump discontinuity at  $t = 3$ .

Graph of step function for rental car cost showing jump discontinuities

Removable Discontinuity:
- Occurs when $\lim_{x \rightarrow a} f(x)$ exists, but $f(a)$ is either undefined or $f(a) \neq \lim_{x \rightarrow a} f(x)$ .
- Example: Piecewise function:

$f(x) = \begin{cases} \frac{x^2 + x - 2}{x - 1} & \text{if } x \neq 1 \\ 4 & \text{if } x = 1 \end{cases}$

* Limit evaluation:  $\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} \frac{(x+2)(x-1)}{x-1} = 3$ .
* Defined value:  $f(1) = 4$ .
* Since  $\lim_{x \rightarrow 1} f(x) = 3 \neq 4 = f(1)$ , a removable discontinuity (hole) exists at  $x = 1$

Graph of function with removable discontinuity and redefined point at x=1

Infinite Discontinuity:
- Occurs when $f(x)$ approaches $\pm\infty$ as $x \rightarrow a$ , forming a vertical asymptote.
- Example: $f(x) = \frac{5x}{x-3}$ at $x = 3$ .
Graphical Assessment Example ( $q(x)$ ):
- $x = -8$ : Removable discontinuity ( $\lim_{x \rightarrow -8} q(x) = -7 \neq -3 = q(-8)$ ).
- $x = -2$ : Infinite discontinuity ( $\lim_{x \rightarrow -2} q(x) = \text{DNE}$ due to asymptote).
- $x = 6$ : Jump discontinuity ( $\lim_{x \rightarrow 6^-} q(x) = 3 \neq 6 = \lim_{x \rightarrow 6^+} q(x)$ ).

The Difference Quotient and the Formal Derivative

Difference Quotient:
- Expresses the average rate of change over a generalized increment $h$ (where $h = \Delta x$ ):

$\text{Difference Quotient} = \frac{f(a + h) - f(a)}{h}$

Graphically represents the slope of the secant line through $(a, f(a))$ and $(a + h, f(a + h))$ .

Diagram showing secant line and difference quotient parameters a and h

Definition of the Derivative at a Point:
- The derivative of $f$ at $x = a$ , denoted $f'(a)$ (read as "f prime of a"), is the limit of the difference quotient as $h \rightarrow 0$ :

$f'(a) = \lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}$

Yields the exact instantaneous rate of change / tangent slope at $x = a$

Algebraic Derivations of Derivatives

Derivation Example 1: Basic Quadratic Function
- Find $f'(1)$ for $f(x) = x^2$
- Form difference quotient at $a = 1$ :

$\frac{f(1 + h) - f(1)}{h} = \frac{(1 + h)^2 - 1^2}{h} = \frac{1 + 2h + h^2 - 1}{h} = \frac{2h + h^2}{h}$

Factor out $h$ (assuming $h \neq 0$ ):

$\frac{h(2 + h)}{h} = 2 + h$

Take limit as $h \rightarrow 0$ :

$f'(1) = \lim_{h \rightarrow 0} (2 + h) = 2 + 0 = 2$

Derivation Example 2: Revenue Function Instantaneous Rate
- Evaluate $R'(20)$ for $R(t) = -0.02t^2 + 2t + 21$ at $t = 20$
- Compute $R(20 + h)$ :

$R(20 + h) = -0.02(20 + h)^2 + 2(20 + h) + 21$ $= -0.02(400 + 40h + h^2) + 40 + 2h + 21$ $= -8 - 0.8h - 0.02h^2 + 40 + 2h + 21 = -0.02h^2 + 1.2h + 53$

Substitute into difference quotient ( $R(20) = 53$ ):

$\frac{R(20 + h) - R(20)}{h} = \frac{(-0.02h^2 + 1.2h + 53) - 53}{h} = \frac{-0.02h^2 + 1.2h}{h}$ $= \frac{h(-0.02h + 1.2)}{h} = -0.02h + 1.2$

Take limit as $h \rightarrow 0$ :

$R'(20) = \lim_{h \rightarrow 0} (-0.02h + 1.2) = 1.2$

Conclusion: Revenue grows at an exact rate of $\$1.2$ million per week at week $20$

The Derivative Function

Formula for the Derivative Function:
- Replacing fixed constant $a$ with variable input $x$ produces a general functional rule $f'(x)$ :

$f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$

Algebraic Derivation Example: $f(x) = x^2 + 4$
- Construct difference quotient:

$\frac{f(x + h) - f(x)}{h} = \frac{[(x + h)^2 + 4] - [x^2 + 4]}{h}$ $= \frac{x^2 + 2xh + h^2 + 4 - x^2 - 4}{h} = \frac{2xh + h^2}{h} = 2x + h$

Take limit as $h \rightarrow 0$ :

$f'(x) = \lim_{h \rightarrow 0} (2x + h) = 2x$

Evaluate at individual points via substitution:
- $f'(1) = 2(1) = 2$
- $f'(-3) = 2(-3) = -6$
- $f'(0) = 2(0) = 0$

Differentiability and Nondifferentiable Functions

Definition of Differentiability:
- A function is differentiable at $x = a$ if its first derivative $f'(a)$ exists.
Relationship Between Continuity and Differentiability:
- Continuity at $x = a$ is a necessary but not sufficient condition for differentiability.
- Discontinuity at $x = a$ guarantees nondifferentiability at $x = a$
- A function can be continuous at a point but fail to be differentiable there.
Causes of Nondifferentiability in Continuous Functions:
- Sharp Corners:
- Occurs when left-hand and right-hand secant slopes approach different finite values.
- Example: Absolute value function $f(x) = |x|$ at $x = 0$ . Slope to left is $-1$ , slope to right is $1$ . No unique tangent line exists.

Graph of absolute value function with corner at x=0

Cusps:
- Occurs where a curve turns infinitely sharply.
- Example: $g(x) = (x - 2)^{2/3}$ at $x = 2$

Graph showing a cusp where derivative is undefined

Vertical Tangents:
- Occurs when secant slopes approach $\pm\infty$ , producing a vertical tangent line.
- Example: $h(x) = (x - 1)^{1/3}$ at $x = 1$ . Tangent line is vertical, so its slope is undefined.

Graph showing vertical tangent line where derivative is undefined

Graphical Nondifferentiability Example:
- A function is non-differentiable at:
- $x = -1$ due to a corner.
- $x = 1$ due to a discontinuity.
- $x = 3$ due to a discontinuity.

Graph evaluating non-differentiability at corners and discontinuities