Comprehensive Calculus I Study Guide: Step-by-Step Guidance

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Q1. Solve the equation: $\ln x + \ln(x-3) = 0$

Background

Topic: Exponential and Logarithmic Equations

This question tests your understanding of logarithmic properties and how to solve equations involving natural logarithms.

Key Terms and Formulas

Natural logarithm: $\ln a$ is the logarithm base $e$ of $a$.
Logarithm property: $\ln a + \ln b = \ln(ab)$
To solve $\ln y = 0$, recall that $e^0 = 1$, so $y = 1$.

Step-by-Step Guidance

Combine the two logarithms using the property $\ln a + \ln b = \ln(ab)$.
Set the resulting logarithmic expression equal to $0$ and rewrite it in exponential form.
Solve the resulting equation for $x$.
Check that your solution(s) make the original logarithms defined (i.e., arguments must be positive).

Try solving on your own before revealing the answer!

Final Answer: $x = 3 + \sqrt{10}$

Combining the logs: $\ln x + \ln(x-3) = \ln[x(x-3)] = 0$ so $x(x-3) = 1$. Solving $x^2 - 3x - 1 = 0$ gives $x = \frac{3 \pm \sqrt{13}}{2}$. Only the value where $x > 3$ is valid, so $x = \frac{3 + \sqrt{13}}{2}$.

Q2. Find the exact value: $\log_{15} 9 + \log_{15} 25$

Background

Topic: Logarithmic Properties and Evaluation

This question tests your ability to use logarithmic addition properties and evaluate logarithms with different arguments.

Key Terms and Formulas

Logarithm addition: $\log_b a + \log_b c = \log_b (ac)$
Change of base and evaluating logarithms

Step-by-Step Guidance

Use the property $\log_b a + \log_b c = \log_b (ac)$ to combine the two logarithms.
Multiply the arguments: $9 \times 25$.
Express the result as a single logarithm.
Try to simplify the argument if possible (e.g., if it is a power of the base).

Try solving on your own before revealing the answer!

Final Answer: $2$

$\log_{15} 9 + \log_{15} 25 = \log_{15} (225) = \log_{15} (15^2) = 2$.

Q3. Compute the limit: $\displaystyle \lim_{x \to 0} \frac{\sin(5x)}{2x}$

Background

Topic: Limits and Trigonometric Functions

This question tests your understanding of limits involving trigonometric functions, especially the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

Key Terms and Formulas

Standard limit: $\lim_{x \to 0} \frac{\sin x}{x} = 1$
Algebraic manipulation to match the standard form

Step-by-Step Guidance

Rewrite the numerator $\sin(5x)$ in terms of $x$.
Factor or multiply numerator and denominator to create the standard $\frac{\sin u}{u}$ form.
Let $u = 5x$ and substitute accordingly.
Apply the standard limit result to evaluate the limit.

Try solving on your own before revealing the answer!

Final Answer: $\dfrac{5}{2}$

Let $u = 5x$, so as $x \to 0$, $u \to 0$. The limit becomes $\lim_{u \to 0} \frac{\sin u}{2(u/5)} = \frac{5}{2} \lim_{u \to 0} \frac{\sin u}{u} = \frac{5}{2}$.

Q4. Find the values of $a$ and $b$ so that the function $f(x) = \begin{cases} x^2 - a & x \leq 1 \\ \frac{3x^2 + 12x - b}{x^2 + 2x - 3} & x > 1 \end{cases}$ is continuous on $(-\infty, \infty)$.

Background

Topic: Continuity of Piecewise Functions

This question tests your understanding of how to ensure a piecewise function is continuous everywhere by matching function values at the point where the definition changes.

Key Terms and Formulas

Continuity at a point: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$
For piecewise functions, set the left and right limits equal at the transition point.

Step-by-Step Guidance

Identify the point where the function changes definition ($x = 1$).
Compute $\lim_{x \to 1^-} f(x)$ using the first piece.
Compute $\lim_{x \to 1^+} f(x)$ using the second piece. Simplify the denominator and numerator if possible.
Set the two limits equal and solve for $a$ and $b$.

Try solving on your own before revealing the answer!

Final Answer: $a = 1$, $b = 14$

For continuity at $x=1$, set $1^2 - a = \lim_{x \to 1^+} \frac{3x^2 + 12x - b}{x^2 + 2x - 3}$. Factoring denominator: $x^2 + 2x - 3 = (x-1)(x+3)$. After simplification and plugging $x=1$, you find $a=1$ and $b=14$.

Q5. Using the Intermediate Value Theorem, determine on which of the intervals (0,1), (1,2), (2,3), (3,4) the equation $x^3 - 3x = 5$ has a root.

Background

Topic: Intermediate Value Theorem (IVT)

This question tests your understanding of the IVT, which states that if a function is continuous on $[a, b]$ and takes values of opposite sign at $a$ and $b$, then it must cross zero somewhere in $(a, b)$.

Key Terms and Formulas

Intermediate Value Theorem: If $f$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then $f$ has a root in $(a, b)$.
Check $f(a)$ and $f(b)$ for each interval.

Step-by-Step Guidance

Define $f(x) = x^3 - 3x - 5$.
Evaluate $f(x)$ at the endpoints of each interval: $x=0,1,2,3,4$.
Check the sign of $f(x)$ at each endpoint.
Identify intervals where $f(x)$ changes sign between endpoints.

Try solving on your own before revealing the answer!

Final Answer: The root is in the interval (1,2)

Evaluating $f(x)$ at the endpoints, you find that $f(1)$ and $f(2)$ have opposite signs, so by the IVT, there is a root in $(1,2)$.

Q6. Compute the derivative of $f(x) = x \cos(x)$ at $x = \frac{\pi}{2}$ and find where the tangent line intersects the y-axis.

Background

Topic: Derivatives and Tangent Lines

This question tests your ability to compute derivatives using the product rule and to find the equation of a tangent line, then determine its y-intercept.

Key Terms and Formulas

Product rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$
Equation of tangent line: $y = f(a) + f'(a)(x - a)$
To find y-intercept, set $x=0$ in the tangent line equation.

Step-by-Step Guidance

Compute $f'(x)$ using the product rule for $f(x) = x \cos(x)$.
Evaluate $f'(x)$ at $x = \frac{\pi}{2}$.
Find $f(\frac{\pi}{2})$.
Write the equation of the tangent line at $x = \frac{\pi}{2}$.
Set $x=0$ in the tangent line equation to find the y-intercept.

Try solving on your own before revealing the answer!

Final Answer: The tangent line intersects the y-axis at $y = -\frac{\pi}{2}$

After computing the derivative and evaluating at $x=\frac{\pi}{2}$, the tangent line equation is $y = f(\frac{\pi}{2}) + f'(\frac{\pi}{2})(x - \frac{\pi}{2})$. Setting $x=0$ gives the y-intercept $y = -\frac{\pi}{2}$.