Calculus Review: Exponential Models, Logarithmic Transformations, Limits, and Continuity

Continuous Interest Rate Models

• Continuous Compounding Model Formula:

  • The formula governing continuous compound interest is:     $A(t) = P \, e^{r t}$
  • Key variables defined:
  • $A(t)$: Account balance at time $t$
  • $P$: Principal amount initially invested
  • $r$: Continuous annual interest rate (as a decimal)
  • $t$: Time elapsed in years

• Continuous Interest Rate Calculation:

  • Initial principal amount: $P = \$25000$
  • Age at investment start: $16$ years old
  • Current age: $23$ years old
  • Elapsed time: $t = 23 - 16 = 7\,\text{years}$
  • Current account balance: $A(7) = \$34250$
  • Step-by-step mathematical determination of interest rate $r$:
  • Substitute values into continuous compounding equation:       $34250 = 25000 \, e^{7 r}$
  • Isolate the exponential term by dividing by $25000$:       $\frac{34250}{25000} = e^{7 r}$$1.37 = e^{7 r}$
  • Take the natural logarithm ($\ln$) of both sides:       $\ln(1.37) = 7 r$
  • Solve for $r$:       $r = \frac{\ln(1.37)}{7} \approx \frac{0.3148107}{7} \approx 0.0449728$
  • Final Interest Rate: $r \approx 4.50\%$

Radioactive Decay and Exponential Models

• Radioactive Decay Principles:

  • Exponential decay is represented by the formula:     $N(t) = N_0 \, e^{-k t}$
  • Alternative formulation using half-life $t_{1/2}$:     $N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}$
  • Decay constant $k$ relationship to half-life $t_{1/2}$:     $k = \frac{\ln(2)}{t_{1/2}}$

• Carbon-14 Dating Problem:

  • Carbon-14 half-life: $t_{1/2} = 5730\,\text{years}$
  • Remaining proportion of original Carbon-14: $43\% = 0.43 N_0$
  • Decay constant calculation:     $k = \frac{\ln(2)}{5730} \approx 0.000120968\,\text{years}^{-1}$
  • Step-by-step determination of bone age $t$:
  • Set up decay equation:       $0.43 N_0 = N_0 \, e^{-k t}$
  • Divide by $N_0$:       $0.43 = e^{-k t}$
  • Apply natural logarithm to both sides:       $\ln(0.43) = -k t$
  • Solve for elapsed time $t$:       $t = \frac{-\ln(0.43)}{k} = \frac{-5730 \times \ln(0.43)}{\ln(2)}$$t \approx \frac{-5730 \times (-0.843970)}{0.693147} \approx 6976.22\,\text{years}$
  • Final Result: The fossilized bone is approximately $6976.22\,\text{years}$ old.

Logarithmic Transformations and Linearization

• Principles of Linearization:

  • Non-linear relationships (power functions and exponential functions) can be transformed into straight-line equations of the form $Y = m X + b$ using logarithmic transformations.

• Power Law Transformation:

  • Equation: $y = 3 x^5$
  • Transformation process applying natural logarithms:     $\ln(y) = \ln(3 x^5)$$\ln(y) = \ln(3) + \ln(x^5)$$\ln(y) = 5 \ln(x) + \ln(3)$
  • Equivalent Linear Form:
  • Let $Y = \ln(y)$ and $X = \ln(x)$$Y = 5 X + \ln(3)$
  • Plot Type: Log-log plot (since logarithmic transformations are applied to both the independent variable $x$ and dependent variable $y$).
  • Linear Components:
  • Slope $m = 5$
  • Vertical intercept $b = \ln(3)$

• Exponential Transformation:

  • Equation: $y = 5 (3^x)$
  • Transformation process applying natural logarithms:     $\ln(y) = \ln(5 \cdot 3^x)$$\ln(y) = \ln(5) + \ln(3^x)$$\ln(y) = \ln(3) x + \ln(5)$
  • Equivalent Linear Form:
  • Let $Y = \ln(y)$$Y = \ln(3) x + \ln(5)$
  • Plot Type: Log-linear plot (semi-log plot; logarithmic transformation applied only to the dependent variable $y$, while $x$ remains untransformed).
  • Linear Components:
  • Slope $m = \ln(3)$
  • Vertical intercept $b = \ln(5)$

Graphical Analysis of Limits, Continuity, and Function Values

• Limit Evaluations from Graph:

  • $\lim_{x \to -4} f(x) = 3$ (Approaching $x = -4$ from both left and right directions targets the open circle at $y = 3$).
  • $\lim_{x \to -2^+} f(x) = \infty$ (As $x$ approaches $-2$ from the right, the function increases without bound along a vertical asymptote).
  • $\lim_{x \to 0} f(x) = \text{DNE}$ (Left-hand limit is $\lim_{x \to 0^-} f(x) = 2$ and right-hand limit is $\lim_{x \to 0^+} f(x) = 1$; since one-sided limits are unequal, the two-sided limit does not exist).
  • $\lim_{x \to 4^-} f(x) = -\infty$ (As $x$ approaches $4$ from the left, the curve drops infinitely downwards along a vertical asymptote).
  • $\lim_{x \to 4^+} f(x) = \infty$ (As $x$ approaches $4$ from the right, the curve decreases from positive infinity along a vertical asymptote).
  • $\lim_{x \to \infty} f(x) = 2$ (As $x$ approaches positive infinity, the function levels off horizontally at $y = 2$).

• Function Value Evaluations from Graph:

  • $f(-4) = 1$ (Represented by the solid point at $(-4, 1)$).
  • $f(-2) = -1$ (Represented by the solid point at $(-2, -1)$).
  • $f(0) = 1$ (Represented by the solid point at $(0, 1)$).
  • $f(4) = \text{Undefined (DNE)}$ (Vertical asymptote at $x = 4$ with no point defined).

• Continuity Analysis at Specific Points:

  • At $x = -4$: Not continuous. There is a removable discontinuity because $\lim_{x \to -4} f(x) = 3$, but $f(-4) = 1 \neq 3$
  • At $x = -3$: Continuous. The line segment passes uninterrupted through $x = -3$, where $\lim_{x \to -3} f(x) = f(-3) = 1.5$
  • At $x = 0$: Not continuous. There is a jump discontinuity because $\lim_{x \to 0^-} f(x) = 2 \neq \lim_{x \to 0^+} f(x) = 1$

• End Behavior as $x \to -\infty$:

  • $\lim_{x \to -\infty} f(x) = \infty$ (The limit does not exist as a finite real number because the function grows endlessly upwards along a line with negative slope).

Evaluation of Algebraic Limits

• Analytical Techniques for Limits:

  • Direct Substitution: Substitute the target $x$-value into the function. If finite, the resulting value is the limit.
  • Indeterminate Form ($\frac{0}{0}$): Requires algebraic simplification (factoring, canceling common factors, or rationalizing with conjugates) before re-evaluating.
  • Infinite Form ($\frac{c}{0}$ for $c eq 0$): Indicates a vertical asymptote; evaluate one-sided limits to determine whether the behavior goes to $+\infty$, $-\infty$, or does not exist.
  • Limits at Infinity: Divide numerator and denominator by the highest power of $x$ in the denominator.

• Detailed Limit Computations:

  • Limit (a):     $\lim_{x \to -6} \frac{x^2 + 3x - 18}{x^2 + 6x}$

  • Substitution yields $\frac{0}{0}$ (Indeterminate).

  • Factor numerator and denominator:       $\frac{x^2 + 3x - 18}{x^2 + 6x} = \frac{(x + 6)(x - 3)}{x(x + 6)}$

  • Cancel $(x + 6)$ for $x \neq -6$:       $\lim_{x \to -6} \frac{x - 3}{x} = \frac{-6 - 3}{-6} = \frac{-9}{-6} = \frac{3}{2}$

  • Limit (b):     $\lim_{x \to \infty} \frac{x^2 + 3x - 18}{x^2 + 6x}$

  • Divide numerator and denominator by $x^2$:       $\lim_{x \to \infty} \frac{1 + \frac{3}{x} - \frac{18}{x^2}}{1 + \frac{6}{x}} = \frac{1 + 0 - 0}{1 + 0} = 1$

  • Limit (c):     $\lim_{x \to -3} \frac{x^2 + 7x + 2}{x + 3}$

  • Direct substitution gives numerator $(-3)^2 + 7(-3) + 2 = -10$ and denominator $-3 + 3 = 0$.

  • Left-hand limit ($x \to -3^-$): $\frac{-10}{\text{small negative}} \to +\infty$

  • Right-hand limit ($x \to -3^+$): $\frac{-10}{\text{small positive}} \to -\infty$

  • Result: One-sided limits differ, so the limit does not exist (DNE).

  • Limit (d):     $\lim_{x \to \infty} \frac{x^2 + 7x + 2}{x + 3}$

  • Numerator degree (2) exceeds denominator degree (1).

  • Divide numerator and denominator by $x$:       $\lim_{x \to \infty} \frac{x + 7 + \frac{2}{x}}{1 + \frac{3}{x}} = \frac{\infty + 7 + 0}{1 + 0} = \infty$

  • Result: Limit does not exist (diverges to $+\infty$).

  • Limit (e):     $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$

  • Direct substitution yields $\frac{0}{0}$ (Indeterminate).

  • Multiply numerator and denominator by conjugate $(\sqrt{x} + 2)$:       $\frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{(x - 4)(\sqrt{x} + 2)} = \frac{x - 4}{(x - 4)(\sqrt{x} + 2)} = \frac{1}{\sqrt{x} + 2}$

  • Evaluate limit:       $\lim_{x \to 4} \frac{1}{\sqrt{x} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{4}$

  • Limit (f):     $\lim_{x \to -\infty} \frac{\sqrt{x} - 2}{x - 4}$

  • Analysis of domain: $\sqrt{x}$ is undefined for real numbers when $x < 0$.

  • As $x \to -\infty$, the function is undefined over the real numbers.

  • Result: Limit does not exist (DNE).

  • Limit (g):     $\lim_{x \to 9^-} \frac{\sqrt{x} + 3}{x - 9}$

  • Substitution yields numerator $\sqrt{9} + 3 = 6$ and denominator $9 - 9 = 0$.

  • As $x \to 9^-$ ($x < 9$), denominator $x - 9$ is a small negative number.

  • Result:       $\lim_{x \to 9^-} \frac{\sqrt{x} + 3}{x - 9} = -\infty$

Continuity Analysis of Rational and Piecewise Functions

• Mathematical Definition of Continuity:

  • A function $f(x)$ is continuous at a point $x = a$ if and only if all three conditions are satisfied:
  1. $f(a)$ is defined.
  2. $\lim_{x \to a} f(x)$ exists.
  3. $\lim_{x \to a} f(x) = f(a)$.

• Continuity Testing of Rational Functions:

  • Function $f(x) = \frac{x^2 + 3x - 18}{x^2 + 6x}$ at $x = -6$:

  • $f(-6) = \frac{0}{0}$, which is undefined.

  • Since $f(-6)$ is undefined, condition 1 fails.

  • Conclusion: $f(x)$ is not continuous at $x = -6$ (it contains a removable discontinuity/hole).

  • Function $f(x) = \frac{x^2 + 7x + 2}{x + 3}$ at $x = 3$:

  • $f(3) = \frac{3^2 + 7(3) + 2}{3 + 3} = \frac{9 + 21 + 2}{6} = \frac{32}{6} = \frac{16}{3}$

  • $\lim_{x \to 3} f(x) = \frac{16}{3}$

  • Since $f(3) = \lim_{x \to 3} f(x) = \frac{16}{3}$, all three conditions are met.

  • Conclusion: $f(x)$ is continuous at $x = 3$

• Solving Parameter $c$ for Continuity in Piecewise Functions:

  • Piecewise definition:     $f(x) = \begin{cases} \frac{2x^2 + 12x - 5}{x + 5} & \text{if } x \le -2 \\ c x + 1 & \text{if } -2 \le x \le 5 \\ 4x + e^{x^2 - 3x - 10} & \text{if } 5 \le x \end{cases}$

  • Boundary 1 Continuity ($x = -2$):

  • Left-hand limit as $x \to -2^-$:       $\lim_{x \to -2^-} f(x) = \frac{2(-2)^2 + 12(-2) - 5}{-2 + 5} = \frac{2(4) - 24 - 5}{3} = \frac{8 - 29}{3} = \frac{-21}{3} = -7$

  • Right-hand limit as $x \to -2^+$:       $\lim_{x \to -2^+} f(x) = c(-2) + 1 = -2c + 1$

  • Equate left-hand and right-hand limits for continuity:       $-2c + 1 = -7 \implies -2c = -8 \implies c = 4$

  • Boundary 2 Continuity Verification ($x = 5$) with $c = 4$:

  • Left-hand limit as $x \to 5^-$:       $\lim_{x \to 5^-} f(x) = 4(5) + 1 = 20 + 1 = 21$

  • Right-hand limit as $x \to 5^+$:       $\lim_{x \to 5^+} f(x) = 4(5) + e^{5^2 - 3(5) - 10} = 20 + e^{25 - 15 - 10} = 20 + e^0 = 20 + 1 = 21$

  • Function value $f(5) = 4(5) + 1 = 21$

  • Since $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5) = 21$, continuity is preserved at $x = 5$

  • Final Solution: $c = 4$