Learn & Review: Precalculus Final Exam

Jan 23, 2026

Precalculus Final Exam Review

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Precalculus Final Exam Review Summary

This video provides a review of key precalculus topics, focusing on exponential equations, inverse functions, logarithms, piecewise functions, polynomial zeros, domain and range, composite functions, and graphing.

1. Solving Exponential Equations

  • Concept: To solve exponential equations, make the bases the same and then set the exponents equal to each other.
  • Example: For $8^{3x+5} = 16^{2x-3}$:
    • Convert bases to 2: $(2^3)^{3x+5} = (2^4)^{2x-3}$
    • Simplify exponents: $2^{9x+15} = 2^{8x-12}$
    • Set exponents equal: $9x + 15 = 8x - 12$
    • Solve for x: $x = -27$

2. Finding Inverse Functions

  • Concept: To find the inverse function $f^{-1}(x)$:
    1. Replace $f(x)$ with $y$.
    2. Swap $x$ and $y$.
    3. Solve for $y$.
    4. Replace $y$ with $f^{-1}(x)$.
  • Alternative Method: To find $f^{-1}(a)$, you can set $f(x) = a$ and solve for $x$.
  • Example: For $f(x) = \frac{8}{x-3}$, find $f^{-1}(4)$:
    • Let $y = \frac{8}{x-3}$.
    • Swap: $x = \frac{8}{y-3}$.
    • Solve for $y$:
      • $x(y-3) = 8$
      • $y-3 = \frac{8}{x}$
      • $y = \frac{8}{x} + 3$
    • So, $f^{-1}(x) = \frac{8}{x} + 3$.
    • Evaluate $f^{-1}(4)$: $f^{-1}(4) = \frac{8}{4} + 3 = 2 + 3 = 5$.
    • Check: $f(5) = \frac{8}{5-3} = \frac{8}{2} = 4$.

3. Evaluating Logarithmic Expressions

  • Concept: The expression $\log_b a = c$ means $b^c = a$.
  • Mental Calculation:
    • $\log_b a$ asks: "To what power must we raise $b$ to get $a$?"
    • If the argument is a fraction (e.g., $\frac{1}{a}$), the exponent is negative.
    • If the base and argument are swapped (e.g., $\log_a b$ instead of $\log_b a$), the exponent is the reciprocal.
  • Algebraic Method:
    1. Set the expression equal to $x$: $\log_b a = x$.
    2. Convert to exponential form: $b^x = a$.
    3. Solve for $x$ by finding a common base.
  • Change of Base Formula: $\log_b a = \frac{\log a}{\log b}$ (using any base, typically base 10 or natural log).
  • Example: Evaluate $\log_{27} \left(\frac{1}{3}\right)$:
    • Let $x = \log_{27} \left(\frac{1}{3}\right)$.
    • Exponential form: $27^x = \frac{1}{3}$.
    • Common base is 3: $(3^3)^x = 3^{-1}$.
    • Simplify: $3^{3x} = 3^{-1}$.
    • Set exponents equal: $3x = -1$.
    • Solve for x: $x = -\frac{1}{3}$.

4. Evaluating Piecewise Functions

  • Concept: A piecewise function is defined by different rules for different intervals of the input. To evaluate it, determine which interval the input value falls into and use the corresponding rule.
  • Example: For $f(x) = \begin{cases} x^2 - 5x + 3 & \text{if } x < 2 \ 4x - 7 & \text{if } x \ge 2 \end{cases}$, find $f(-2) + f(3)$:
    • $f(-2)$: Since $-2 < 2$, use the first rule: $(-2)^2 - 5(-2) + 3 = 4 + 10 + 3 = 17$.
    • $f(3)$: Since $3 \ge 2$, use the second rule: $4(3) - 7 = 12 - 7 = 5$.
    • Sum: $17 + 5 = 22$.

5. Solving Logarithmic Equations

  • Concept: Convert the logarithmic equation to its equivalent exponential form.
  • Formula: $\log_b a = c \iff b^c = a$.
  • Example: Solve $\log_3 (8x+1) = 4$:
    • Exponential form: $3^4 = 8x + 1$.
    • Calculate $3^4$: $81 = 8x + 1$.
    • Solve for x:
      • $80 = 8x$
      • $x = 10$.

6. Continuous Compounding (PERT Formula)

  • Formula: $A = Pe^{rt}$
    • $A$: Future value
    • $P$: Principal amount
    • $e$: Euler's number (approx. 2.718)
    • $r$: Annual interest rate (as a decimal)
    • $t$: Time in years
  • Concept: Used when interest is compounded continuously. To solve for time ($t$), use natural logarithms.
  • Example: Karen invests $15,000 at 9.5% compounded continuously. How long to double?
    • $P = 15000$, $A = 30000$, $r = 0.095$.
    • $30000 = 15000 e^{0.095t}$
    • $2 = e^{0.095t}$
    • Take natural log of both sides: $\ln 2 = \ln(e^{0.095t})$
    • $\ln 2 = 0.095t \cdot \ln e$ (since $\ln e = 1$)
    • $t = \frac{\ln 2}{0.095} \approx 7.3$ years.

7. Compound Interest (General Formula)

  • Formula: $A = P\left(1 + \frac{r}{n}\right)^{nt}$
    • $A$: Future value
    • $P$: Principal amount
    • $r$: Annual interest rate (as a decimal)
    • $n$: Number of times interest is compounded per year
    • $t$: Time in years
  • Concept: Used for interest compounded a specific number of times per year (e.g., monthly, quarterly).
  • Example: $30,000 at 9% compounded monthly for 40 years.
    • $P = 30000$, $r = 0.09$, $n = 12$, $t = 40$.
    • $A = 30000\left(1 + \frac{0.09}{12}\right)^{12 \times 40}$
    • $A = 30000(1 + 0.0075)^{480}$
    • $A = 30000(1.0075)^{480} \approx $1,083,297.06$.

8. Finding Zeros of Polynomials

  • Concept: Zeros are the x-values where the polynomial equals zero ($f(x)=0$).
  • Method:
    1. List possible rational zeros using the Rational Root Theorem (factors of the constant term divided by factors of the leading coefficient).
    2. Test possible zeros by plugging them into the polynomial until one results in $f(x)=0$.
    3. Use synthetic division with the found zero to reduce the polynomial's degree.
    4. Factor the resulting polynomial (or use the quadratic formula if it's a quadratic) to find the remaining zeros.
    5. Sum all the zeros.
  • Example: Find the sum of zeros for $f(x) = x^3 - 3x^2 - 10x + 24$.
    • Possible rational zeros: $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$.
    • Test $x=2$: $f(2) = 2^3 - 3(2^2) - 10(2) + 24 = 8 - 12 - 20 + 24 = 0$. So, $x=2$ is a zero.
    • Synthetic division with 2: 
      2 | 1  -3  -10   24
  |    2   -2  -24
  ----------------
    1  -1  -12    0
    • The remaining polynomial is $x^2 - x - 12$.
    • Factor the quadratic: $(x-4)(x+3)$.
    • The zeros are $x=2$, $x=4$, and $x=-3$.
    • Sum of zeros: $2 + 4 + (-3) = 3$.

9. Domain of Radical Functions

  • Concept: The expression inside a square root (or any even root) must be non-negative (greater than or equal to zero).
  • Method: Set the expression inside the radical $\ge 0$ and solve the inequality.
  • Example: Find the domain of $f(x) = \sqrt{5-x}$.
    • Set $5-x \ge 0$.
    • Subtract 5: $-x \ge -5$.
    • Divide by -1 and reverse inequality: $x \le 5$.
    • Domain in interval notation: $(-\infty, 5]$.

10. Domain of Rational Functions

  • Concept: A rational function is undefined when its denominator is zero. These values must be excluded from the domain.
  • Method:
    1. Set the denominator equal to zero.
    2. Solve for $x$. These are the values to exclude.
    3. Write the domain in interval notation, using parentheses for excluded values and infinity.
  • Example: Find the domain of $f(x) = \frac{1}{6x^2 + x - 15}$.
    • Set denominator to zero: $6x^2 + x - 15 = 0$.
    • Factor the quadratic (using factoring by grouping or other methods): $(2x-3)(3x+5) = 0$.
    • Solve for x:
      • $2x - 3 = 0 \implies x = \frac{3}{2}$
      • $3x + 5 = 0 \implies x = -\frac{5}{3}$
    • The values $x = \frac{3}{2}$ and $x = -\frac{5}{3}$ are excluded.
    • Domain in interval notation: $\left(-\infty, -\frac{5}{3}\right) \cup \left(-\frac{5}{3}, \frac{3}{2}\right) \cup \left(\frac{3}{2}, \infty\right)$.

11. Solving Logarithmic Equations

  • Concept: Use logarithm properties to combine terms and then convert to exponential form. Be aware of extraneous solutions because the argument of a logarithm must be positive.
  • Properties: $\log_b M + \log_b N = \log_b (MN)$
  • Example: Solve $\log_2(x+5) + \log_2(x+1) = 5$.
    • Combine logs: $\log_2((x+5)(x+1)) = 5$.
    • Expand: $\log_2(x^2 + 6x + 5) = 5$.
    • Convert to exponential form: $x^2 + 6x + 5 = 2^5$.
    • $x^2 + 6x + 5 = 32$.
    • Set to zero: $x^2 + 6x - 27 = 0$.
    • Factor: $(x+9)(x-3) = 0$.
    • Potential solutions: $x = -9$ and $x = 3$.
    • Check for extraneous solutions:
      • If $x = -9$: $\log_2(-9+5) = \log_2(-4)$ is undefined. Extraneous.
      • If $x = 3$: $\log_2(3+5) = \log_2(8)$ and $\log_2(3+1) = \log_2(4)$. Both are defined.
    • The only valid solution is $x = 3$.

12. Evaluating Composite Functions

  • Concept: To find $f(g(x))$, first evaluate the inner function $g(x)$ at the given value, then use that result as the input for the outer function $f$.
  • Example: If $f(x) = x^2 - 4x + 3$ and $g(x) = 3x + 5$, find $f(g(-1))$.
    • Evaluate $g(-1)$: $g(-1) = 3(-1) + 5 = -3 + 5 = 2$.
    • Evaluate $f(2)$: $f(2) = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$.
    • So, $f(g(-1)) = -1$.

13. Identifying Functions from Graphs (Vertical Line Test)

  • Concept: A graph represents a function if and only if every vertical line drawn intersects the graph at most once.
  • Method: Apply the Vertical Line Test. If any vertical line crosses the graph more than once, it is not a function.
  • Example: Graph C fails the vertical line test because a single vertical line can intersect it at multiple points. Graphs A, B, and D pass the test.

14. Graphing Absolute Value Functions and Finding Domain/Range

  • Concept: Absolute value functions typically form a V-shape. Transformations shift, stretch, compress, or reflect this basic shape.
  • Transformations:
    • $y = |x|$: Parent function (V-shape, vertex at origin).
    • $y = |x| + k$: Shifts vertically up by $k$.
    • $y = |x| - k$: Shifts vertically down by $k$.
    • $y = |x+h|$: Shifts horizontally left by $h$.
    • $y = |x-h|$: Shifts horizontally right by $h$.
    • $y = -|x|$: Reflects across the x-axis (opens downward).
  • Domain: For absolute value functions, the domain is typically all real numbers $(-\infty, \infty)$.
  • Range: Depends on the vertex and whether the graph opens up or down. If it opens up, the range starts at the y-coordinate of the vertex and goes to $\infty$. If it opens down, the range starts at $-\infty$ and goes up to the y-coordinate of the vertex.
  • Example: Graph $y = -|x+2| + 3$.
    • Parent: $y=|x|$.
    • Shift left 2 units: $y=|x+2|$.
    • Reflect across x-axis: $y=-|x+2|$.
    • Shift up 3 units: $y=-|x+2|+3$.
    • Vertex: $(-2, 3)$. Opens downward.
    • Domain: $(-\infty, \infty)$.
    • Range: $(-\infty, 3]$.

15. Graphing Exponential Functions and Finding Domain/Range/Asymptotes

  • Concept: Exponential functions have the form $y = a \cdot b^{x-h} + k$.
  • Key Features:
    • Horizontal Asymptote: $y = k$. This is the line the graph approaches but never touches.
    • Domain: $(-\infty, \infty)$ (unless restricted).
    • Range: Depends on the asymptote and whether the graph is reflected or shifted. If $b>1$ and not reflected, range is $(k, \infty)$. If reflected, range is $(-\infty, k)$.
  • Graphing Method:
    1. Identify the horizontal asymptote ($y=k$).
    2. Choose convenient x-values relative to the horizontal shift ($x-h$). Often setting $x-h=0$ and $x-h=1$ is useful.
    3. Calculate the corresponding y-values.
    4. Plot the points and sketch the curve, approaching the asymptote.
  • Example: Graph $y = 2^{x+3} + 1$.
    • Horizontal Asymptote: $y = 1$.
    • Let $x+3 = 0 \implies x = -3$. $y = 2^0 + 1 = 1 + 1 = 2$. Point: $(-3, 2)$.
    • Let $x+3 = 1 \implies x = -2$. $y = 2^1 + 1 = 2 + 1 = 3$. Point: $(-2, 3)$.
    • Plot $(-3, 2)$ and $(-2, 3)$. Draw the curve starting near $y=1$ and passing through these points, extending to the right.
    • Domain: $(-\infty, \infty)$.
    • Range: $(1, \infty)$ (since the graph is above the asymptote $y=1$).

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