MAT 271 Final Exam Review Notes

Part I Problem Set 1

• The final exam for MAT 271 covers Chapters 2 and 3 in the first part and Chapters 4 and 5 in the second part.
• The best way to prepare is by re-working problems.

Problem 1

• Given the graph of $g(t)$. We evaluate the following:
  • a) $g(5) = 1$
  • b) $\lim_{t \to 5} g(t) = -1$
  • c) $g(3) = -1$
  • d) $\lim_{t \to 3^-} g(t) = -1$
  • e) $\lim_{t \to 3^+} g(t) = -1$
  • f) $\lim_{t \to 3} g(t) = -1$
  • g) $g(2)$ is undefined.
  • h) $\lim_{t \to 2} g(t) = -1.5$
  • i) $g(-2) = 5$
  • j) $\lim_{t \to -2^-} g(t) = 5$
  • k) $\lim_{t \to -2^+} g(t) = -3.5$
  • l) $\lim_{t \to -2} g(t)$ does not exist (DNE).

Problem 2

• Find the value of the constant $a$ that makes the function continuous on $(-\infty, \infty)$.
• $f(x) = \begin{cases} x + 2a & \text{if } x < 1 \ ax^2 + 5 & \text{if } x \geq 1 \end{cases}$
• $a = 4$

Problem 3

• Use the limit definition of the derivative to find $f'(x)$ for $f(x) = 2x^2 + x$.
• The limit definition of the derivative is: $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
• $f'(x) = 4x + 1$

Problem 4

• Find $y'$ for $y = \arctan(x^3 - 1)$. Do not simplify.
• $y' = \frac{3x^2}{1 + (x^3 - 1)^2}$

Problem 5

• Find $y'$ for $y = x^2 \tan(\frac{1}{x})$. Do not simplify.
• $y' = \frac{-\sec^2(\frac{1}{x})}{x^2} + 2x \tan(\frac{1}{x})$

Problem 6

• Find $\frac{dy}{dx}$ by implicit differentiation and then evaluate the derivative at the point $(0,0)$.
• $\sin(x^2 + y) = x^2 + 2x + 3y$
• $y' = \frac{2x + 2 - 2x\cos(x^2 + y)}{\cos(x^2 + y) - 3}$
• At the point $(0,0)$, $y' = -1$

Problem 7

• Let $f(x) = xe^x$. Find the local minimum value(s), local maximum value(s), absolute minimum value, and absolute maximum value.
• Local minimum: $(-1, e^{-1})$
• Local maximum: none
• Absolute Minimum: $(-1, e^{-1})$
• Absolute Maximum: none

Problem 8

• Find the limit:
• $\lim_{x \to \infty} \frac{x+2}{9x^2+1} = 0$

Problem 9

• Find the limit:
• $\lim_{x \to 0} \frac{\sqrt{3+x} - \sqrt{3}}{x} = \frac{1}{2\sqrt{3}}$

Problem 10

• Find the limit:
• $\lim_{x \to 4} \frac{x^2 - 5x + 4}{x^2 - 2x - 8} = \frac{1}{6}$

Part I Problem Set 2

Problem 1

• Given the graph of $f(x)$. We evaluate the following:
  • a) $\lim_{x \to -2^-} f(x) = 3$
  • b) $\lim_{x \to -2^+} f(x) = 3$
  • c) $\lim_{x \to -2} f(x) = 3$
  • d) $f(-2) = 6$
  • e) $\lim_{x \to 1^-} f(x) = -3$
  • f) $\lim_{x \to 1^+} f(x) = -1$
  • g) $\lim_{x \to 1} f(1)$ DNE (Does Not Exist)
  • h) $f(1) = -1$
  • i) $f(-4)$ Undefined

Problem 2

• Find $\lim<em>{x \to 0} \frac{|x|}{x}$. $\lim</em>{x \to 0} \frac{|x|}{x}$ does not exist.

Problem 3

• Find $f'(x)$. Do not simplify.
• $f(x) = x^7 - 2x^3 + e^2$
• $f'(x) = 7x^6 - 6x^2$

Problem 4

• Find $f'(x)$. Do not simplify.
• $f(x) = \frac{x + e^{\pi}}{\cos^4 + \sin^5(6x)}$
• $f'(x) = \frac{(\cos^4 + \sin^5(6x)) - (x + e^{\pi})(30\sin^4(6x) \cdot \cos(6x))}{(\cos^4 + \sin^5(6x))^2}$

Problem 5

• Use logarithmic differentiation to find $f'(x)$ for $f(x) = (\sin(x))^x$. DO NOT SIMPLIFY.
• $y' = (\sin x)^x \left( \frac{x \cos x}{\sin x} + \ln(\sin x) \right)$

Problem 6

• A stone dropped in a pond sends out a circular ripple whose radius increases at a constant rate of $4 \text{ ft/s}$. After 12 seconds, how rapidly is the area enclosed by the ripple increasing?
• $\frac{dA}{dt} = 384\pi \text{ ft}^2/\text{sec} \approx 1206.4 \text{ ft}^2/\text{sec}$

Problem 7

• Given $f(x) = 3x^4 - 4x^3 + 2012$. Use calculus to answer the following:
  • a) Find the intervals where $f$ is increasing and decreasing.
    • Decreasing: $(-\infty, 1)$
    • Increasing: $(1, \infty)$
  • b) Find all local and absolute minimum and maximum values and the values of $x$ at which they occur.
    • Local Minimum: $(1, f(1)) = (1, 2011)$
    • Local Maximum: none
    • Absolute Minimum: $(1, f(1)) = (1, 2011)$
    • Absolute Maximum: none
  • c) Find the intervals where $f$ is concave up and concave down.
    • Concave Up: $(-\infty, 0) \cup (\frac{2}{3}, \infty)$
    • Concave Down: $(0, \frac{2}{3})$
  • d) Find all points of inflection (Round to thousandths).
    • Inflection points: $(0, f(0)) = (0, 2012)$, $(\frac{2}{3}, f(\frac{2}{3})) = (\frac{2}{3}, 2011.407)$