Rutgers Calculus I Quiz 2 Review – Step-by-Step Guidance

Q1. For which values of $a$ is the following function continuous?

$f(x) = \begin{cases} x^2 + 8x + 12 & x \neq -6 \\ a & x = -6 \end{cases}$

Background

Topic: Continuity of Piecewise Functions

This question tests your understanding of continuity at a point, especially for piecewise-defined functions. You need to ensure the function is continuous everywhere, particularly at $x = -6$ where the definition changes.

Key Terms and Formulas

• Continuity at a point $c$: $\lim_{x \to c} f(x) = f(c)$

• Limit from the left and right: Both must exist and be equal to $f(c)$ for continuity.

Step-by-Step Guidance

1. Check continuity at all points except $x = -6$. For $x \neq -6$, $f(x)$ is a polynomial, which is continuous everywhere.

2. Focus on $x = -6$, where the function definition changes. For continuity at $x = -6$, you need $\lim_{x \to -6} f(x) = f(-6)$.

3. Compute $\lim_{x \to -6} f(x)$ using the first case: $x^2 + 8x + 12$.

4. Set $\lim_{x \to -6} f(x)$ equal to $a$ (since $f(-6) = a$) and solve for $a$.

Try solving on your own before revealing the answer!

Q2. Use the Intermediate Value Theorem to show that $\sqrt{3x^2 + 7x + 5} = 3$ has a solution in $(0, 1)$. Verify the hypotheses first.

Background

Topic: Intermediate Value Theorem (IVT)

This question tests your ability to apply the IVT to show that an equation has a solution in a given interval. You must check the hypotheses of the theorem before applying it.

Key Terms and Formulas

• Intermediate Value Theorem: If $f$ is continuous on $[a, b]$ and $N$ is between $f(a)$ and $f(b)$, then there exists $c$ in $(a, b)$ such that $f(c) = N$.

• Define $f(x) = \sqrt{3x^2 + 7x + 5}$ and consider $f(x) = 3$.

Step-by-Step Guidance

1. Define $f(x) = \sqrt{3x^2 + 7x + 5}$ and rewrite the equation as $f(x) = 3$.

2. Check that $f(x)$ is continuous on $[0, 1]$ (the interval containing $(0, 1)$).

3. Evaluate $f(0)$ and $f(1)$ to see if $3$ lies between these values.

4. If $3$is between$f(0)$and$f(1)$, by IVT, there exists$c$in$(0, 1)$such that$f(c) = 3$.

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Q3. Use the limit definition of the derivative to compute the derivative of $f(x) = 3x^2 + 2x - 1$.

Background

Topic: Definition of the Derivative

This question tests your understanding of the formal (limit) definition of the derivative and your ability to apply it to a polynomial function.

Key Terms and Formulas

• Limit definition of the derivative:

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

Step-by-Step Guidance

1. Write the limit definition for $f'(x)$ using $f(x) = 3x^2 + 2x - 1$.

2. Compute $f(x + h)$ by substituting $x + h$ into the function.

3. Subtract $f(x)$ from $f(x + h)$ and simplify the numerator.

4. Divide the simplified numerator by $h$.

5. Set up the limit as $h \to 0$ (do not evaluate yet).

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Q4. Compute the first derivative for the following functions using the rules of differentiation.

Background

Topic: Differentiation Rules

This question tests your ability to apply various differentiation rules, such as the power rule, product rule, quotient rule, and chain rule, to different types of functions.

Key Terms and Formulas

• Power Rule: $\frac{d}{dx} x^n = n x^{n-1}$

• Product Rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$

• Quotient Rule: $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

• Chain Rule: $\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$

• Derivatives of trig functions: $\frac{d}{dx} \cos x = -\sin x$, $\frac{d}{dx} \sec x = \sec x \tan x$

Step-by-Step Guidance

(a) $f(x) = (3\sqrt{x})^7$

1. Rewrite $3\sqrt{x}$ as $3x^{1/2}$, so $f(x) = (3x^{1/2})^7$.

2. Expand to $f(x) = 3^7 x^{7/2}$.

3. Apply the power rule to differentiate $f(x)$.

(b) $f(x) = 7x^{11} - 9x^7$

1. Apply the power rule to each term separately.

(c) $f(x) = \cos x (7x^2 - x^{7/2})$

1. Recognize this as a product of two functions: $u(x) = \cos x$, $v(x) = 7x^2 - x^{7/2}$.

2. Apply the product rule: $f'(x) = u'(x)v(x) + u(x)v'(x)$.

3. Compute $u'(x)$ and $v'(x)$ using the derivative rules.

(d) $f(x) = \frac{9x^2 - 7x + 5}{x^3 + 6}$

1. Identify numerator $u(x) = 9x^2 - 7x + 5$ and denominator $v(x) = x^3 + 6$.

2. Apply the quotient rule.

3. Compute $u'(x)$ and $v'(x)$.

(e) $f(x) = \cos x \sec x$

1. Recall that $\sec x = 1/\cos x$, so $f(x) = 1$ for all $x$ where $\cos x \neq 0$.

2. Differentiate $f(x) = 1$.

(f) $f(x) = \frac{\sqrt{x}}{4} - 9x^5 - e^{x^2}$

1. Rewrite $\sqrt{x}$ as $x^{1/2}$.

2. Differentiate each term separately using the power rule and the chain rule for $e^{x^2}$.

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Q5. Find the equation of the tangent line to $y = x^3 - 4x^2 + 2x - 1$ at $x = 2$.

Background

Topic: Tangent Lines and Derivatives

This question tests your ability to find the equation of a tangent line to a curve at a given point using derivatives.

Key Terms and Formulas

• Derivative: The slope of the tangent line at a point $x = a$ is $f'(a)$.

• Point-slope form: $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the point of tangency.

Step-by-Step Guidance

1. Find $f'(x)$ for $f(x) = x^3 - 4x^2 + 2x - 1$ using the power rule.

2. Evaluate $f'(2)$ to get the slope of the tangent line at $x = 2$.

3. Find the $y$-coordinate at $x = 2$ by computing $f(2)$.

4. Write the equation of the tangent line using the point-slope form with $x_1 = 2$ and $y_1 = f(2)$.

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Q6. Is the function $f(x) = \frac{x^2 - 4x + 3}{x - 3}$ continuous at $a = 3$? Use the continuity checklist.

Background

Topic: Continuity at a Point

This question tests your ability to check continuity at a specific point using the three-part continuity checklist.

Key Terms and Formulas

• Continuity Checklist at $x = a$:

  1. $f(a)$ is defined.

  2. $\lim_{x \to a} f(x)$ exists.

  3. $\lim_{x \to a} f(x) = f(a)$.

Step-by-Step Guidance

1. Check if $f(3)$ is defined by substituting $x = 3$ into the function.

2. Compute $\lim_{x \to 3} f(x)$ by simplifying the expression if possible.

3. Compare the value of the limit to $f(3)$ to determine continuity.

Try solving on your own before revealing the answer!