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

Average Rate of Change and Instantaneous Rate of Change

• Conceptual Distinction Between Rates of Change:

  • Average Rate of Change: Measures the overall change in output per unit change in input over a specific interval. It indicates how fast or slow a function $y = f(x)$ is increasing or decreasing on average between two points $x = x_1$ and $x = x_2$.
  • Instantaneous Rate of Change: Measures the exact rate at which a function is changing at one specific instant in time.
  • Distance and Speed Metaphor:
    • If a vehicle travels $200\,\text{miles}$ in $4\,\text{hours}$, the average rate of change (average speed) is:             $\text{Average Speed} = \frac{200\,\text{miles}}{4\,\text{hours}} = 50\,\text{mph}$
    • If the driver looks at the speedometer exactly $1\,\text{hour}$ into the trip and reads $62\,\text{mph}$, that measurement represents the instantaneous rate of change at that specific instant.

• Mathematical Definition of Average Rate of Change:

  • The average rate of change of a function $f(x)$ with respect to $x$, as $x$ changes from $x_1$ to $x_2$, is the ratio of the change in the output $f(x)$ to the change in the input $x$:         $\text{Average Rate of Change} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

• Graphical Interpretation:

  • Graphically, the average rate of change of $f(x)$ between $x_1$ and $x_2$ is equal to the slope of the secant line passing through the points $(x_1, f(x_1))$ and $(x_2, f(x_2))$ on the graph of $y = f(x)$.

The Difference Quotient

• Purpose and Formulation:

  • The difference quotient is an algebraic process and notation developed to easily manipulate and evaluate the average rate of change of a function.
  • Instead of using two distinct input symbols $x_1$ and $x_2$, set $x_1 = x$ and $x_2 = x + h$, where $h$ represents a non-zero change in $x$ ($h \neq 0$).

• Algebraic Derivation:

  • Substituting $x_1 = x$ and $x_2 = x + h$ into the average rate of change ratio yields:         $\text{Average Rate of Change} = \frac{f(x_2) - f(x_1)}{x_2 - x_1} = \frac{f(x + h) - f(x)}{(x + h) - x} = \frac{f(x + h) - f(x)}{h}$
  • Constraint: $h \neq 0$.

• Graphical Connection:

  • The difference quotient still represents the slope of a secant line connecting the points $(x, f(x))$ and $(x + h, f(x + h))$.

Applications of Average Rate of Change

• Lunar Motion Example:

  • A stone is thrown upward from the surface of the moon with an initial velocity of $3.2\,\text{m/s}$. Its position in meters above the surface after $t$ seconds is modeled by:         $f(t) = 3.2t - 0.8t^2$
  • Part A: Calculating Average Rate of Change from $t = 1\,\text{second}$ to $t = 2\,\text{seconds}$:
    • Evaluate position at $t = 1\,\text{s}$:             $f(1) = 3.2(1) - 0.8(1)^2 = 3.2 - 0.8 = 2.4\,\text{m}$
    • Evaluate position at $t = 2\,\text{s}$:             $f(2) = 3.2(2) - 0.8(2)^2 = 6.4 - 3.2 = 3.2\,\text{m}$
    • Calculate the average rate of change:             \text{Average Rate of Change} = \frac{f(2) - f(1)}{2 - 1} = \frac{3.2 - 2.4}{1} = 0.8\,\text{m\,s^{-1}}
  • Part B: Graphical Representation:
    • The graph of $f(t)$ forms a downward-opening parabola with a vertex at $(2, 3.2)$.
    • The secant line connects the points $(1, 2.4)$ and $(2, 3.2)$, having a positive slope equal to $0.8$.

• Compound Interest Savings Account Example:

  • An initial investment of $\$2000$ is placed into a savings account paying $6\%$ annual interest compounded quarterly. The future amount $A(t)$ in dollars after $t$ years is given by:         $A(t) = 2000(1.015)^{4t}$(Note: The quarterly growth factor is $1 + \frac{0.06}{4} = 1.015$).
  • Part A: Value and Context of $A(3)$:
    • Calculation:             $A(3) = 2000(1.015)^{4(3)} = 2000(1.015)^{12} \approx 2391.24$
    • Context Interpretation: $A(3) \approx 2391.24\,\text{dollars}$ represents the total balance in the savings account after $3\,\text{years}$.
  • Part B: Value and Context of $A(5) - A(3)$:
    • Calculation for $A(5)$:             $A(5) = 2000(1.015)^{4(5)} = 2000(1.015)^{20} \approx 2693.71$
    • Difference:             $A(5) - A(3) = 2693.71 - 2391.24 = 302.47$
    • Context Interpretation: $A(5) - A(3) = 302.47\,\text{dollars}$ represents the total amount of interest earned (net growth) between year $3$ and year 5$.\n * **Part C: Value and Context of \frac{A(5) - A(3)}{5 - 3}**:\n * Calculation:\n            \frac{A(5) - A(3)}{5 - 3} = \frac{302.47}{2} = 151.235\n * *Context Interpretation*: \frac{A(5) - A(3)}{5 - 3} = 151.235\,\text{dollars\,year^{-1}}$represents the average rate of change (average growth per year) of the money in the savings account over the$2\,\text{year}$period from year$3$to year$5$.

Algebraic Simplification of the Difference Quotient

• Required Function Types:

  • Simplifying and using the difference quotient $\frac{f(x+h)-f(x)}{h}$ applies to linear, quadratic, and simple rational functions.

• Complete Derivation for Quadratic Function $f(x) = 16 - x - 2x^2$:

  • Step 1: Write down the difference quotient expression:         $\frac{f(x + h) - f(x)}{h} = \frac{[16 - (x + h) - 2(x + h)^2] - [16 - x - 2x^2]}{h}$
  • Step 2: Expand $(x + h)^2$ and distribute negative signs:         $\frac{16 - x - h - 2(x^2 + 2xh + h^2) - 16 + x + 2x^2}{h}$
  • Step 3: Distribute the $-2$ into the quadratic expansion:         $\frac{16 - x - h - 2x^2 - 4xh - 2h^2 - 16 + x + 2x^2}{h}$
  • Step 4: Combine like terms in the numerator (terms without $h$ cancel out):         $\frac{-h - 4xh - 2h^2}{h}$
  • Step 5: Factor out $h$ from the numerator:         $\frac{h(-1 - 4x - 2h)}{h}$
  • Step 6: Cancel $h$ in the numerator and denominator:         $-1 - 4x - 2h$

• Numerical Table Evaluation at $x = 1$:

  • For $x = 1$, the simplified difference quotient reduces to:         $-1 - 4(1) - 2h = -5 - 2h$
  • Table of Average Rates of Change as $h \rightarrow 0$:
    • When $x = 1$ and $h = 0.1$:             $\text{Difference Quotient} = -5 - 2(0.1) = -5.2$
    • When $x = 1$ and $h = 0.01$:             $\text{Difference Quotient} = -5 - 2(0.01) = -5.02$
    • When $x = 1$ and $h = 0.001$:             $\text{Difference Quotient} = -5 - 2(0.001) = -5.002$

Intuitive Introduction to Limits

• Concept of a Limit:

  • The limit process involves examining the behavior of a function $f(x)$ as the input $x$ values approach a specific number $c$.
  • The number $c$ may or may not belong to the domain of $f(x)$.
  • Limits are used to determine function behavior at critical points such as holes, gaps, cusps, corners, or vertical asymptotes.
  • Graphical and numerical methods yield educated estimations of limits.

• One-Sided Limits:

  • Left-Sided Limit:
    • If $f(x)$ approaches a value $L$ as $x$ approaches $c$ from the left ($x < c$), it is written as:             $\lim_{x \rightarrow c^-} f(x) = L$
    • Meaning: As values of $x$ are chosen closer and closer to $c$ (while remaining strictly less than $c$), the output values $y = f(x)$ get closer and closer to L$.\n * **Right-Sided Limit**:\n * If f(x)$approaches a value$M$as$x$approaches$c$from the right ($x > c), it is written as:\n            \lim_{x \rightarrow c^+} f(x) = M\n * *Meaning*: As values of x$are chosen closer and closer to$c$(while remaining strictly greater than$c$), the output values$y = f(x)$get closer and closer to$M$.

• Existence Theorem for Two-Sided Limits:

  • Let $f$ be a function defined for all values of $x$ near $x = c$, with the possible exception of $c$ itself.
  • Theorem: The general two-sided limit exists if and only if both one-sided limits exist and are equal to the same value $L$:         $\lim_{x \rightarrow c^-} f(x) = \lim_{x \rightarrow c^+} f(x) = L \iff \lim_{x \rightarrow c} f(x) = L$

Numerical Estimation of Limits

• Table Estimation Example:

  • Evaluating the limit:         $\lim_{x \rightarrow 2} \frac{x^2 + 4x - 12}{x - 2}$
  • Left-Hand Table ($x < 2$):
    • $x = 1.99 \implies f(1.99) = \frac{(1.99)^2 + 4(1.99) - 12}{1.99 - 2} = 7.99$
    • $x = 1.999 \implies f(1.999) = 7.999$
    • $x = 1.9999 \implies f(1.9999) = 7.9999$
    • $x = 1.99999 \implies f(1.99999) = 7.99999$
  • Right-Hand Table ($x > 2$):
    • $x = 2.01 \implies f(2.01) = \frac{(2.01)^2 + 4(2.01) - 12}{2.01 - 2} = 8.01$
    • $x = 2.001 \implies f(2.001) = 8.001$
    • $x = 2.0001 \implies f(2.0001) = 8.0001$
    • $x = 2.00001 \implies f(2.00001) = 8.00001$
  • Limit Determination:
    • Directly substituting $x = 2$ into the expression produces $\frac{0}{0}$, which causes a division-by-zero error.
    • As $x$ approaches $2$ from both the left and right, the function values approach $8$.
    • Therefore:             $\lim_{x \rightarrow 2} \frac{x^2 + 4x - 12}{x - 2} = 8$

• Core Principles of Limit Computation:

  1. Function Existence vs. Limit Existence: A function does not need to exist at $x = c$ for the limit at $x = c$ to exist.
  2. Fundamental Question of Limits: Limits ask for the behavior of $y = f(x)$ NEAR the $x$ value as $x$ approaches $c$.
  3. Irrelevance of Value at $x = c$: Computing a limit does not depend on the actual value of $f(c)$. The limit relies exclusively on outputs surrounding $x = c$ (to the immediate left and right).

Limits and Asymptotes

• Limits at Infinity (End Behavior):

  • $\lim_{x \rightarrow \infty} f(x)$ describes the horizontal end behavior of the graph as $x$ grows infinitely positive.
  • $\lim_{x \rightarrow -\infty} f(x)$ describes the horizontal end behavior of the graph as $x$ grows infinitely negative.

• Infinite Limits and Vertical Asymptotes:

  • If $f(x)$ approaches positive or negative infinity as $x$ approaches a finite number $c$, the line $x = c$ is a vertical asymptote.

• Non-Existence of Limits:

  • If left-sided and right-sided limits approach different finite values, or if the function oscillates endlessly, the two-sided limit does not exist (DNE).

Continuity of Functions

• Intuitive Definition of Continuity:

  • A function is continuous at a specific point $x = c$ if its graph contains no holes, gaps, jumps, or breaks at that point.

• Precise Definition of Continuity at a Point:

  • A function $f$ is continuous at a number $x = c$ if and only if the following three conditions are met:
    1. $c$ is in the domain of $f$ (meaning $f(c)$ is defined).
    2. $\lim_{x \rightarrow c} f(x)$ exists.
    3. $\lim_{x \rightarrow c} f(x) = f(c)$.

• Continuity Over an Interval:

  • A function $f$ is continuous over an open interval $(a, b)$ if and only if it is continuous at every individual number within that interval.

• Continuity Theorem for Standard Functions:

  • Linear, quadratic, polynomial, rational, power, and root functions are continuous over their entire domains and discontinuous at all values not in their domain.
  • To determine where standard algebraic functions are continuous, determine their domain.

Domain and Continuity Determination Examples

• Example A: Polynomial Function:

  • Function: $f(x) = 12x^3 - 2x^2 + 15x - 5$
  • Domain: Polynomials are defined for all real numbers, $(-\infty, \infty)$.
  • Continuity: $f(x)$ is continuous on $(-\infty, \infty)$.

• Example B: Rational Function:

  • Function: $g(x) = \frac{x^2 - 6x + 2}{x + 7}$
  • Domain Condition: The denominator cannot equal zero:         $x + 7 \neq 0 \implies x \neq -7$
  • Domain: $(-\infty, -7) \cup (-7, \infty)$.
  • Continuity: $g(x)$ is continuous on $(-\infty, -7) \cup (-7, \infty)$ and discontinuous at x = -7$.\n\n* **Example C: Root Function**:\n * Function: g(x) = \sqrt{x + 10}\n * Domain Condition: The expression under an even root must be non-negative:\n        x + 10 \ge 0 \implies x \ge -10\n * Domain: [-10, \infty).\n * Continuity: g(x)$is continuous on$[-10, \infty)$$.