Calculus Midterm Study Guidance

Q1. For the function $f(x)$, given in the following figure, find (make reasonable estimates from the grid, if necessary):

Background

Topic: Reading values and limits from a graph.

This question tests your ability to interpret a graph to find function values and limits at specific points.

Key Terms:

• Function value: $f(a)$ is the $y$-value of the function at $x=a$.

• Limit: $\lim_{x \to a} f(x)$ is the value the function approaches as $x$ gets close to $a$.

Step-by-Step Guidance

1. To find $f(5)$, locate $x=5$ on the $x$-axis and read the corresponding $y$-value on the graph.

2. To find $f(-9)$, do the same for $x=-9$.

3. For $\lim_{x \to 7} f(x)$, observe the behavior of $f(x)$ as $x$ approaches $7$ from both sides. Check if the left and right limits agree.

4. For $\lim_{x \to 2^-} f(x)$, look at the value $f(x)$ approaches as $x$ approaches $2$ from the left only.

Try solving on your own before revealing the answer!

Final Answers:

• $f(5) = 3$

• $f(-9) = 7$

• $\lim_{x \to 7} f(x) = 2$

• $\lim_{x \to 2^-} f(x)$ does not exist (DNE)

These values are estimated by reading the graph at the specified points and considering the behavior as $x$ approaches the given values.

Q2. Evaluate the following limits (show supporting work):

Background

Topic: Limits and algebraic manipulation.

This question tests your ability to evaluate limits using substitution, factoring, and algebraic simplification.

Key Terms and Formulas:

• Limit: $\lim_{x \to a} f(x)$

• Factoring: Used to simplify expressions when direct substitution gives an indeterminate form.

• L'Hospital's Rule: If $\lim_{x \to a} \frac{f(x)}{g(x)}$ gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$ (if the limit exists).

Step-by-Step Guidance

1. For each limit, first try direct substitution of the value for $x$.

2. If you get an indeterminate form (like $0/0$), factor or simplify the numerator and denominator.

3. For limits at infinity, divide numerator and denominator by the highest power of $x$ in the denominator.

4. For trigonometric limits, recall that $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

5. Set up the simplified expression for each limit, but stop before plugging in the final value.

Try solving on your own before revealing the answer!

Final Answers:

• (a) $\lim_{x \to 4} \frac{x}{x-4}$ does not exist (infinite limit).

• (b) $\lim_{x \to 3} \frac{x+3}{x^2-6}$ is $\frac{6}{3}$ or $2$.

• (c) $\lim_{x \to \infty} \frac{x+3}{x^2+6}$ is $0$.

• (d) $\lim_{x \to 0} \frac{\sin 3x}{\sin 2x} = \frac{3}{2}$.

Each answer is found by applying the appropriate algebraic or trigonometric technique.

Q3. Find derivatives of the following functions:

Background

Topic: Differentiation rules (sum, product, chain, quotient, and implicit differentiation).

This question tests your ability to apply differentiation rules to various types of functions.

Key Terms and Formulas:

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

• Product Rule: $\frac{d}{dx}[u \cdot v] = u'v + uv'$

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

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

• Derivative of $\sin x$: $\cos x$

• Derivative of $\sec x$: $\sec x \tan x$

• Derivative of $e^{x}$: $e^{x}$

Step-by-Step Guidance

1. For each function, identify which differentiation rule(s) apply (sum, product, chain, or quotient).

2. Apply the rule to write the derivative in terms of $x$ (do not simplify fully yet).

3. For (e), use the definition of the derivative: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.

4. Set up the derivative expressions, but stop before fully simplifying or plugging in values.

Try solving on your own before revealing the answer!

Final Answers:

• (a) $y' = 6x + 2\sec x \tan x$

• (b) $y' = \cos x + 2$

• (c) $y' = 2x \sin(3x) + x^2 \cdot 3\cos(3x)$

• (d) $y' = \frac{e^x (1 + x^2) - e^x \cdot 2x}{(1 + x^2)^2}$

• (e) $y' = \frac{1}{2\sqrt{x}}$

Each derivative is found by applying the appropriate rule and simplifying as needed.

Q4. The position of an object moving horizontally after $t$ seconds is given by $s(t) = 10 + 12t - t^3$, for $t > 0$, where $s(t)$ is measured in feet. Determine the acceleration of the object when its velocity is zero.

Background

Topic: Motion along a line (kinematics), derivatives as velocity and acceleration.

This question tests your understanding of how to find velocity and acceleration from a position function using derivatives.

Key Terms and Formulas:

• Velocity: $v(t) = s'(t)$

• Acceleration: $a(t) = v'(t) = s''(t)$

Step-by-Step Guidance

1. Find the velocity function by differentiating $s(t)$ with respect to $t$.

2. Set the velocity function equal to zero and solve for $t$ to find when the velocity is zero.

3. Find the acceleration function by differentiating the velocity function with respect to $t$.

4. Substitute the value of $t$ (when velocity is zero) into the acceleration function, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer:

The acceleration when velocity is zero is $a = -18$ ft/s$^2$.

This is found by differentiating twice and evaluating at the correct $t$ value.

Q5. For the curve $(y-x)^2 = 1 + x$:

Background

Topic: Implicit differentiation and tangent lines.

This question tests your ability to use implicit differentiation to find $\frac{dy}{dx}$ and to find the equation of a tangent line at a given point.

Key Terms and Formulas:

• Implicit Differentiation: Differentiate both sides with respect to $x$, treating $y$ as a function of $x$.

• Tangent Line: $y - y_1 = m(x - x_1)$, where $m$ is the slope at $(x_1, y_1)$.

Step-by-Step Guidance

1. Differentiate both sides of $(y-x)^2 = 1 + x$ with respect to $x$ using the chain rule and product rule as needed.

2. Solve for $\frac{dy}{dx}$ in terms of $x$ and $y$.

3. To find the tangent line at $(0,1)$, substitute $x=0$ and $y=1$ into your expression for $\frac{dy}{dx}$ to get the slope.

4. Write the equation of the tangent line using the point-slope form, but do not simplify to the final equation yet.

Try solving on your own before revealing the answer!

Final Answers:

• (a) $\frac{dy}{dx} = \frac{1 + 2y - 2x}{2y - 2x}$

• (b) The equation of the tangent line at $(0,1)$ is $y = x + 1$.

Implicit differentiation and substitution at the given point yield the slope and tangent line equation.