Skip to main content
Mathematics LibreTexts

Exam1

  • Page ID
    219447
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\dsum}{\displaystyle\sum\limits} \)

    \( \newcommand{\dint}{\displaystyle\int\limits} \)

    \( \newcommand{\dlim}{\displaystyle\lim\limits} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \(\newcommand{\longvect}{\overrightarrow}\)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

     

     

     

     

    Name                                    

     

     

     

     

     

    MATH 105 PRACTICE EXAM 1

     

     

    Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Show your work.

     

     

     

    Key

     

    PROBLEM 1  Please answer the following true or false.  If false, explain why or provide a counter example.  If true, explain why.

     

     

     

    A)     (7 Points)  If f(x)  and g(x)  are differentiable functions with

               f '(5)  =  10    and    g'(5)  =  4

    then if

                             f(x)
         h(x)   =           -  3g(x)
                          2 

            then

                h'(5) = -7

     

    Solution

     

    B)     (7 Points)  Let f(x)  and g(x)  be continuous functions
         f(1)  >  g(1)       and      f(2) < g(2)
    then if
         h(x) = f(x) - g(x)
    h(x)  has a root for some value of x between 1 and 2.

     

    Solution

     

    C)     (7 Points)  Let f(x)  and g(x)  be continuous functions such that

               \( \lim\limits_{x \to 2} g(x) = 0 \)   and    \( h(x) = \frac{f(x)}{g(x)} \) 

                Then h(x) has a vertical asymptote at x = 2.

     

    Solution

     

    PROBLEM 2 Find the following limits if they exist:

    A) (8 Points) \( \lim\limits_{x \to \frac{3}{4}} \frac{6x^2 +19x - 36}{3x^2 - 7x + 4} \)

     

    Solution

     

    B) (8 Points)  \( \lim\limits_{x \to 0} \frac{x}{tan(x)} \)

     

    Solution

     

    C)    (8 Points)  \( \lim\limits_{x \to 2} \frac{\sqrt{x-2} + 2}{x+2} \)

     

    Solution

     

     

     

     

    PROBLEM 3

     

     graph with a v shape near (-3,0), a hole at (-1,2), a cup at (0,1), an asymptote at x=1, and a jump discontinuity at x = 3 with left to y=1 right to y=3 and the point at (3,2).

     

     

     

    A)   (8 Points)  Find the following limits if they exist

                       i)  \( \lim\limits_{x \to -3^{+}} f(x) \)  ii) \( \lim\limits_{x \to -1} f(x) \)     iii) \( \lim\limits_{x \to 0} f(x) \)     iv) \( \lim\limits_{x \to 1} f(x) \)      v) \( \lim\limits_{x \to 3} f(x) \) 

     

    Solution

     

    B)    (8 Points)  At which values is f(x) not continuous?

     

    Solution

     

    C)   (8 Points)  At which values is f(x) not differentiable?

     

    Solution

     

     

     

     

    PROBLEM 4   (20 Points)  Below is the function y = f(x).  Sketch a graph of the derivative y = f ’(x).

     Graph down until x=-4, then flattens from x=-4 to x=5, then up until x = 0, then a slightly downward slope with a horizontal asymptote on the right side.

     

    Solution

     

     

    PROBLEM 5   Find f ' (x) for the following

    A)    (10 Points)  \(  f(x) = x^2 sec(x) - \frac{2x^3}{cot(x)} \) 

     

    Solution

     

    B)     (11 Points)  \(  f(x) =  \frac{2}{x^5} - (x - cos(x))^2 + \frac{x^3}{\sqrt{x}} + \pi^2 \) 

    Solution

     

     

     

    PROBLEM 6 Let \(  f(x) = \sqrt{x-1} \) 

    A)    (10 Points)  Use the limit definition of the derivative to find f ’(x).

    Solution

     

    B)     (10 Points)  Prove using the \(\epsilon - \delta \)  definition of the limit that \( \lim\limits_{x \to 2} 6 - 2x = 2 \)

    Solution

     

     

    PROBLEM 7

     

    The position of a robin flying through the wind is given by

               s(t)  =  -5t + tcost

    Find its acceleration when t is 2 seconds.

    Solution

     

     

     

     

    Extra Credit:  Write down one thing that your instructor can do to make the class better and one thing that you want to remain the same in the class.

    (Any constructive remark will be worth full credit.)

    Back to Differential Calculus Page

     

     

     

     

     

    Exam1 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?