\input epsf \input graphicx \magnification 1200 \vsize 9.5truein \nopagenumbers \parskip 10pt \parindent 0 pt Exam 2 Nov. 10, 2005 \hfil SHOW ALL WORK \hfil \vskip -10pt Math 25 Calculus I \hfil Either circle your answers or place on answer line. \hfil [16] 1.) If $f'(x) = 3x^4 + 2 + 4x^{-1}$ and $f(1) = 8$, find $f$ $f(x) = {3 \over 5}x^5 + 2x + 4 ln(x) + C$ for some $C$. $f(1) = 8$, $f(1) = 8 = {3 \over 5}(1)^5 + 2(1) + 4 ln(1) + C$ $8 = {3 \over 5} + 2 + 0 + C$ $6 - {3 \over 5} = {30 - 3 \over 5} = {27 \over 5} = C$ \centerline{Answer 1.) $\underline{~f(x) = {3 \over 5}x^5 + 2x + 4 ln(x) + {27 \over 5} ~}$} \vskip 10pt [16] 2.)$lim_{x \rightarrow 0^+} [x^{x^3}]$~= $\underline{~1~}$ $lim_{x \rightarrow 0^+} [x^{x^3}]$~= $lim_{x \rightarrow 0^+} [e^{ln(x^{x^3})}]$~= $lim_{x \rightarrow 0^+} [e^{x^3ln(x)}]$ $lim_{x \rightarrow 0^+} [{x^3ln(x)}]$ = $lim_{x \rightarrow 0^+} [{ln(x) \over x^{-3}}]$ = $lim_{x \rightarrow 0^+} [{{1 \over x} \over -3x^{-4}}]$ = $lim_{x \rightarrow 0^+} [{x^3 \over -3}] = 0$ Hence $lim_{x \rightarrow 0^+} [e^{x^3ln(x)}]$ = $e^{lim_{x \rightarrow 0^+} [{x^3ln(x)}]}$ = $e^0$ = 1 \vskip 10pt [10] 3a.) Given $x^2 + 2xy + y^3 - x - 3 = 0$, then $y'= \underline{~{1 - 2x - 2y \over 2x + 3y^2} ~}$ ${d \over dx}(x^2 + 2xy + y^3 - x - 3) = {d \over dx}(0)$, $2x + 2(y + xy') + 3y^2y' - 1 = 0$, $2x + 2y + 2xy' + 3y^2y' - 1 = 0$, $(2x + 3y^2)y' = 1 - 2x - 2y$, $y' = {1 - 2x - 2y \over 2x + 3y^2} $, \vskip 10pt [6] 3b.) Find the equation of the tangent line to the curve $x^2 + 2xy + y^3 - x - 3 = 0$, at the point (-2, 1). $y' = {1 - 2x - 2y \over 2x + 3y^2} $, Hence at (-2, 1), $y' = {1 - 2(-2) - 2(1) \over 2(-2) + 3(1)^2} = {1 + 4 - 2 \over -4 + 3} = -3 $. ${y - 1 \over x - (-2)} = -3$, $ y = -3(x + 2) + 1 = -3x - 6 + 1 = -3x - 5$ \vskip 10pt \centerline{Answer 3b.) $\underline{~y = -3x - 5~}$} \vskip 10pt [16] 4.) If $g(3) = 4$ and $g'(x) \leq 2$, how large can $g(8)$ be? $\underline{~14~}$ Since $g'$ exists, $g$ is differentiable and hence continuous. By the MVT since $g$ continuous on [3, 8] and $g$ differentiable on (3, 8), then there exists $c \in [3, 8]$ such that $${g(8) - g(3) \over 8 - 3} = g'(c)$$ Hence ${g(8) - 4 \over 5} = g'(c) \leq 2$. Thus $g(8) \leq 2(5) + 4 = 14$. [16] 5.) A tank is in the form of an inverted cone having a height of 16m and a diameter at the top of 8 m. Water is flowing into the tank at the rate of 2 m$^3$/min. How fast is the water level rising when the water is 5m deep? (Volume of cone = ${1 \over 3}\pi r^2 h$) ${dV \over dt} = 2$, ${dh \over dt} = $? when $h = 5$. NOTE: V, h, r change with respect to time. None of them are constant. $V = {1 \over 3}\pi r^2 h$, ~~${r \over h} = { 4 \over 16}$. Hence $r = {1 \over 4}h$. $V = {1 \over 3}\pi ({1 \over 4}h)^2 h = {1 \over 3}\pi ({h^3 \over 16}) $ ${d \over dt}[V] = {d \over dt}[{1 \over 3}\pi ({h^3 \over 16}) ]$ ${dV \over dt} = {1 \over 3}\pi ({3h^2 \over 16}{dh \over dt}) $ ${dV \over dt} = \pi ({h^2 \over 16}{dh \over dt}) $ $2 = \pi ({5^2 \over 16}{dh \over dt}) $. Hence ${32 \over 25\pi} = {dh \over dt} $ Alternate answer: $V = {1 \over 3}\pi r^2 h$. NOTE: V, h, r change with respect to time. ${d \over dt}[V] = {d \over dt}[{1 \over 3}\pi r^2 h]$ ${dV \over dt} = {1 \over 3}\pi [2r{dr \over dt}h + r^2{dh \over dt}]$ When $h = 5$: $2 = {1 \over 3}\pi [2r{dr \over dt}(5) + r^2{dh \over dt}]$ ${r \over h} = { 4 \over 16}$. Hence $r = {1 \over 4}h$. Thus ${dr \over dt} = {1 \over 4}{dh \over dt} $ and when $h = 5$, $r = {5 \over 4} $. Thus, when $h = 5$: $2 = {1 \over 3}\pi [2({5 \over 4})({1 \over 4}){dh \over dt} (5) + ({5 \over 4})^2{dh \over dt}] = {25 \pi \over 16}{dh \over dt}$. Hence ${32 \over 25\pi} = {dh \over dt} $ \vfill \centerline{Answer 5.) $\underline{~{dh \over dt} = {32 \over 25\pi} ~}$} \eject 6.) Find the following for $f(x) = {x \over (x-1)^2}$ (if they exist; if they don't exist, state so). Use this information to graph $f$. Note $f'(x) = {-x-1 \over (x-1)^3}$ and $f''(x) = {2(x+2) \over (x-1)^4}$ %%Is $f$ even, odd, periodic? What is the domain and range of $f$? \vskip -2pt [1] 6a.) critical numbers: $\underline{~-1~}$ [1.5] 6b.) local maximum(s) occur at $x = \underline{~none~}$ [1.5] 6c.) local minimum(s) occur at $x = \underline{~-1~}$ [1.5] 6d.) The global maximum of $f$ on the interval [0, 5] is $\underline{~none~}$ and occurs at $x = \underline{~none~}$ [1.5] 6e.) The global minimum of $f$ on the interval [0, 5] is $\underline{~0~}$ and occurs at $x = \underline{~0~}$ [1.5] 6f.) Inflection point(s) occur at $x = \underline{~-2~}$ [1] 6g.) $f$ increasing on the intervals $\underline{~(-1, 1)~}$ [1] 6h.) $f$ decreasing on the intervals $\underline{~(-\infty, -1) \cup (1, \infty)~}$ [1.5] 6i.) $f$ is concave up on the intervals $\underline{~(-2,1) \cup (1, \infty)~ }$ [1.5] 6j.) $f$ is concave down on the intervals$\underline{~(-\infty, -2)~ }$ [1.5] 6k.) Equation(s) of vertical asymptote(s)$\underline{~x = 1~}$ [4] 6l.) Equation(s) of horizontal and/or slant asymptote(s)$\underline{~y = 0~}$ \vskip -4pt [4] 6m.) Graph $f$ \vskip -8pt \includegraphics[width=48ex]{grapht} \end \end{document} 3.7--4.) Find $f'''(x) 6.) 3.11) 1 or 23, 27, 31 9.) 4.9 7, 3, 8, 5 3.8--2.) If $f(x) = [sin(x)]^2$, then $f'(x) = \underline{\hskip 4in}$ Simplify answer \vskip 10pt [15] 2.) Find the derivative of $f(x) = \sqrt{x}$ by using the definition of derivative. \vfill \centerline{Answer 2.) $\underline{\hskip 4in}$} \eject Find the following derivatives \vskip 10pt [15]~ 3.) ${d \over dx}[{e^x(x^2 - x + 3) \over cos(2x)}]$ \vfil \centerline{Answer 3.) $\underline{\hskip 4in}$} \vskip 10pt [15]~ 4.) ${d \over dx}[2sin(e^{x^3} + 4)]$ \vfil \centerline{Answer 4.) $\underline{\hskip 4in}$} \eject 5.) Answer the following questions based on the graph of $f$ given below. \centerline{\includegraphics[width=36ex]{graph25a}} %%\centerline{\epsfbox{graph25a.eps}} [2] 5a.) domain of $f = \underline{\hskip 0.7in}$ \hfil [2] 5b.) range of $f = \underline{\hskip 0.7in}$ \vskip 15pt [1] 5c.) Is $f$ one-to-one? $\underline{\hskip 0.7in}$ \hfil [2] 5d.) Does $f^{-1}$ exist? $\underline{\hskip 0.7in}$ \vskip 15pt [1] 5e.) $f(1) = \underline{\hskip 0.7in}$ \hfil [2] 5g.) $f\hskip 1pt'(-1) = \underline{\hskip 0.7in}$ \vskip 15pt [2] 5f.) Solve $f(x) = 1: \underline{\hskip 0.7in}$ \vskip 15pt [2] 5i.) $lim_{x \rightarrow +\infty} f(x) = \underline{\hskip 0.7in}$ \hfil [2] 5j.) $lim_{x \rightarrow -\infty} f(x) = \underline{\hskip 0.7in}$ \vskip 15pt [2] 5k.) $lim_{x \rightarrow 3^+} f(x) = \underline{\hskip 0.7in}$ \hfil [2] 5l.) $lim_{x \rightarrow 3^-} f(x) = \underline{\hskip 0.7in}$ \vskip 15pt [2] 5m.) State all points where $f$ is not continuous: $\underline{\hskip 1.7in}$ \vskip 15pt [2] 5n.) State all points where $f$ is not differentiable: $\underline{\hskip 1.7in}$ %%\centerline{\epsfbox{graph25.eps,width=6truein}} \includegraphics[width=83ex]{graph25} \end 6.) Given the graph of $y = f(x)$ below, draw the following graphs: [4] 6a) y = \hfil \end Review Problems Ch 3 p.270-274 Concept check: 1, 3 TF Quiz: 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 13 Exercises: 1 - 14, 17, 19, 21-24, 26, 30, 32, 35, 37, 39, 41, 42, 44, 55-57, 59, 61, 63, 67, 68, 84, [85-87 (but ignore acceleration)], 88, 100, 101, 102 Ch 2 p. 176 - 179 Concept check: 2, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14 TF Quiz: All Exercises: [1, 2], 3-12, [13], 14, [15-21], 25, 26, [31, 32], [35], 37, [39], 40, 41ab, [44-46], [47ab], 49 Ch 1: Concept check: 1, 3, 4, 6, 7, 8, 9, 11, 12, 13 Exercises: 1, 2, 5-8, 10, 19, 24, 25, 26 Tuesday Quiz: Proof + 3.6 Prove $(f + g)' = f' + g' $ $(cf)' = c(f')$ $f$ differentiable implies $f$ continuous: Show $lim_{x \rightarrow a} f(x) = f(a)$ Show $lim_{x \rightarrow a} [f(x) - f(a)] = 0$ $f$ differentiable at $a$ implies $f'(a)$ exists and $lim_{x \rightarrow a} {f(x) - f(a) \over x - a} = f'(a)$ Hence $lim_{x \rightarrow a} [f(x) - f(a)] = lim_{x \rightarrow a} {f(x) - f(a) \over x - a} (x-a) = lim_{x \rightarrow a} f'(a) (x - a) f'(a) lim_{x \rightarrow a} (x - a) = f'(a) (0) = 0$. $lim_{x \rightarrow a} f(x) = lim_{x \rightarrow a} [f(x) - f(a) + f(a)] = lim_{x \rightarrow a} [f(x) - f(a)] + lim_{x \rightarrow a} f(a) = 0 + f(a)$ \end