NOTE: You may view the original PDFs of the notes written in class by clicking on the link for the 1:30 Section or 2:30 Section, respectively.

Midterm 1 Review (covering Lectures 2-10)

  1. Problem 1.Use the table below to compute numerically \( \lim_{x \to 0} \frac{e^{3x} - 1}{x} \):

    Values of \( f(x) \) near \( x = 0 \)
    \( x \) -0.1 -0.01 -0.001 0 0.001 0.01 0.1
    \( f(x) \) 2.5918 2.9534 2.9955 — 3.0045 3.0455 3.4986

    For \( f(x) = \frac{e^{3x} - 1}{x} \), we compute \( f(-0.001) \) by pushing the following buttons on the calculator in this order:

    1. 0.001
    2. +/-
    3. X 3 =
    4. 2ND LN
    5. − 1
    6. =
    7. ÷ 0.001
    8. +/-
    9. =

    Repeating this for all values in the table, we see \( f(x) \to 3 \) as \( x \to 0 \).

    So,

    \[ \lim_{x \to 0} \frac{e^{3x} - 1}{x} = 3 \]

  2. Problem 2.Evaluate:

    \[ \lim_{x \to 0} \frac{\sin(x) - x}{4x} \]

    We use algebra to rewrite this in a way we can compute the limit

    \[ \begin{aligned} \lim_{x \to 0} \frac{\sin(x) - x}{4x} &= \frac{1}{4} \lim_{x \to 0} \frac{\sin(x) - x}{x} \\ &= \frac{1}{4} \lim_{x \to 0} \left( \frac{\sin x}{x} - \frac{x}{x} \right) \\ &= \frac{1}{4} \lim_{x \to 0} \left( \frac{\sin x}{x} - 1 \right) \\ &= \frac{1}{4} (1 - 1) \\ &= \frac{1}{4} \cdot 0 \\ &= 0 \end{aligned} \]

  1. Problem 3.Locate and classify all discontinuities of the function \( f(x) \) below:

    \[ f(x) = \begin{cases} -1 & x < -1 \\ \frac{x^2 + x}{2x} & -1 \le x \le 1 \\ -\cos(\pi x) & x > 1 \end{cases} \]

    We start by checking for discontinuitres everywhere except \( x = -1 \) and \( x = 1 \).

    1. \( -1 \) is continuous, so \( f \) is continuous on \( (-\infty, -1) \)

    2. \( \frac{x^2 + x}{2x} = \frac{x(x+1)}{x \cdot 2} \) is continuous everywhere except \( 0 \). Moreover, the limit as \( x \to 0 \) exists:

      \[ \lim_{x \to 0} \frac{x(x+1)}{x\,\cdot 2} = \lim_{x \to 0} \frac{(x+1)}{2} = \frac{1}{2} \]

      Therefore, \( f \) is continuous on \( (-1,0) \cup (0,1) \) and has a hole at \( x = 0 \).

    3. \( -\cos(\pi x) \) is continuous, so \( f \) is continuous on \( (1, \infty) \).

    We now check for continuity at \( x = -1 \):

    \[ \begin{aligned} \lim_{x \to -1^-} f(x) &= \lim_{x \to -1^-} -1 = -1 \\ \lim_{x \to -1^+} f(x) &= \lim_{x \to -1^+} \frac{x^2 + x}{2x} = \frac{(-1)^2 + (-1)}{2(-1)} = \frac{0}{-2} = 0 \\ f(-1) &= 0 \end{aligned} \]

    So, the left and right limits are unequal implying \( f \) has a jump discontinuity at \( x = -1 \).

We now check for continuity at \( x = 1 \):

\[ \begin{aligned} \lim_{x \to 1^-} f(x) &= \lim_{x \to 1^-} \frac{x^2 + x}{2x} = \frac{(1)^2 + 1}{2(1)} = 1 \\ \lim_{x \to 1^+} f(x) &= \lim_{x \to 1^+} -\cos(\pi x) = -\cos(\pi) = -(-1) = 1 \\ f(1) &= 1 \end{aligned} \]

Therefore, \( f \) is continuous at \( x = 1 \).

  1. Problem 4.Use the definition of the derivative to compute \( f'(x) \) if \( f(x) = x^2 - x \).

    By definition,

    \[ \begin{aligned} f'(x) &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{(x+h)^2 - (x+h) - x^2 + x}{h} \\ &= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x - h - x^2 - x}{h} \\ &= \lim_{h \to 0} \frac{2xh + h^2 - h}{h} \\ &= \lim_{h \to 0} (2x + h - 1) \\ &= 2x - 1 \end{aligned} \]

  1. Problem 5.Find the equation of the line tangent to \( f(x) = \sqrt{x} \) at \( x = 4 \)

    To find the slope, we need to compute \( f'(4) \):

    \[ \begin{aligned} f'(x) &= \left[ x^{\frac{1}{2}} \right]' \\ &= \frac{1}{2} x^{-\frac{1}{2}} \\ &= \frac{1}{2\sqrt{x}} \\ f'(4) &= \frac{1}{2\sqrt{4}} \\ &= \frac{1}{4} \end{aligned} \]

    To find the point the tangent line touches the graph, we need to compute \( f(4) \)

    \[ \begin{aligned} f(x) &= \sqrt{x} \\ f(4) &= \sqrt{4} = 2 \end{aligned} \]

    So, the point is \( (4, f(4)) \), i.e., \( (4,2) \).

    Now, by the equation for the tangent line is:

    \[ \begin{aligned} y - f(4) &= f'(4)(x-4) \\ y - 2 &= \frac{1}{4}(x-4) \\ y - 2 &= \frac{1}{4}x - 1 \\ y &= \frac{1}{4}x + 1 \end{aligned} \]

  1. Problem 6.Differentiate the following functions:

    1. \( f(x) = 3x^4 + 7\sqrt[4]{x^3} - 5x^{-\frac{2}{5}} + 9x + 1 \)

      \[ \begin{aligned} f(x) &= 3x^4 + 7x^{\frac{3}{4}} - 5x^{-\frac{2}{5}} + 9x + 1 \\ f'(x) &= 3\cdot 4x^3 + 7\cdot\frac{3}{4}x^{-\frac{1}{4}} - 5\left(-\frac{2}{5}\right)x^{-\frac{7}{5}} + 9\cdot 1 \\ f'(x) &= 12x^3 + \frac{21}{4}x^{-\frac{1}{4}} + 2x^{-\frac{7}{5}} + 9 \end{aligned} \]

      where we used \( [x^p] = px^{p-1} \), \( [f+g]' = f' + g' \), and \( [kf]' = k\cdot f' \)

    2. \( g(x) = e^x \csc(x) \)

      We need the product rule \( [fg]' = f'g + fg' \)

      \[ \begin{aligned} g'(x) &= [e^x]'\csc x + e^x[\csc x] \\ &= e^x \csc x + e^x(-\csc x \cot x) \\ &= e^x \csc x (1 - \cot x) \end{aligned} \]

    3. \( h(x) = \frac{\sec(x)}{\sin(x)} \)

      Here we need the quotient rule \( \left[\frac{f}{g}\right]' = \frac{f'g - fg'}{g^2} \)

      \[ \begin{aligned} h'(x) =& \frac{[\sec x]'\sin x - \sec x[\sin x]'}{\sin^2 x} \\[0.5em] h'(x) =& \frac{\sec x \tan x \sin x - \sec x \cos x}{\sin^2 x}\\[0.5em] =& \sec^2 x - \csc^2 x \end{aligned} \]

    4. \( s(x) = \tan(x^2 + 3x) \)

      Here we need the chain rule \( [f \circ g]' = f'(g(x))\cdot g'(x) \)

      \[ \begin{aligned} s'(x) &= \sec^2(x^2 + 3x)\left[x^2 + 3x\right]' \\ &= \sec^2(x^2 + 3x)\cdot(2x + 3) \\ &= (2x+3)\sec^2(x^2 + 3x) \end{aligned} \]

  1. Problem 7.The position of a dampened pendulum is measured from the angle (in radians) from its resting point. Its position after t seconds is given by the equation:

    \[ s(t) = e^{-t}\cos\left(\frac{\pi}{2}t\right) \]

    1. What is the velocity function \( v(t) \) (in radians per second)?

      Recall that the velocity is the derivative of the position function:

      \[ \begin{aligned} v(t) = s'(t) &= \left[e^{-t}\right]'\cos\left(\frac{\pi}{2}t\right) + e^{-t}\left[\cos\left(\frac{\pi}{2}t\right)\right]' \\ &= -e^{-t}\cos\left(\frac{\pi}{2}t\right) + e^{-t}\left[-\frac{\pi}{2}\sin\left(\frac{\pi}{2}t\right)\right] \\ &= -e^{-t}\left[\cos\left(\frac{\pi}{2}t\right) + \frac{\pi}{2}\sin\left(\frac{\pi}{2}t\right)\right]\frac{\text{rad}}{\text{s}} \end{aligned} \]

    2. What is the velocity after 1 second?

      \[ \begin{aligned} v(1) &= -e^{-1}\left[\cos\left(\frac{\pi}{2}\right) + \frac{\pi}{2}\sin\left(\frac{\pi}{2}\right)\right] \\ &= -e^{-1}\left[0 + \frac{\pi}{2}\right] \\ &= -\frac{\pi}{2e} \\ &= \left[-\frac{\pi}{2e}\right]\frac{\text{rad}}{\text{s}} \end{aligned} \]

  1. Problem 8.The limit below represents the derivative of some function \( f(x) \) at a point \( a \):

    \[ \lim_{h \to 0} \frac{\sqrt[3]{\frac{1}{64+h}} - \frac{1}{4}}{h} \]

    1. Find the function \( f(x) \) and the point \( a \).

      We rewrite the limit in the form \( \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h} \):

      \[ \begin{aligned} \lim_{h \to 0} \frac{\sqrt[3]{\frac{1}{64+h}} - \frac{1}{4}}{h} &= \lim_{h \to 0} \frac{\sqrt[3]{(64+h)^{-1}} - \frac{1}{4}}{h} \\ &= \lim_{h \to 0} \frac{(64+h)^{-\frac{1}{3}} - \frac{1}{64^{1/3}}}{h} \\ &= \lim_{h \to 0} \frac{(64+h)^{-\frac{1}{3}} - (64)^{-\frac{1}{3}}}{h} \end{aligned} \]

      Thus, \( f(x) = x^{-\frac{1}{3}} \) and \( a = 64 \).

    2. Compute the limit.

      \[ \begin{aligned} \lim_{h \to 0} \frac{\sqrt[3]{\frac{1}{64+h}} - \frac{1}{4}}{h} &= \left. \frac{d}{dx}\left(x^{-\frac{1}{3}}\right) \right|_{x=64} \\ &= \left. -\frac{1}{3} x^{-\frac{4}{3}} \right|_{x=64} \\ &= \left. -\frac{1}{3}\cdot\frac{1}{\left(\sqrt[3]{x}\right)^{4}} \right|_{x=64} \\ &= -\frac{1}{3\left(\sqrt[3]{64}\right)^{4}} \\ &= -\frac{1}{3(\sqrt[3]{64})^{4}} \\ &= -\frac{1}{3\cdot 256} \\ &= -\frac{1}{768} \end{aligned} \]