Practice exam 1

(a) The form. At $x=0$ the numerator is $e^{0}-1=0$ and the denominator is $\ln1=0{:}$ numerator and denominator both tend to $0$ as $x\to0.$ The quotient is of the form $\tfrac00$ (Definition 5.18), whatever $b$ is, so l'Hôpital's rule (Fact 5.19) may be used.

(a) Differentiate numerator and denominator separately. By the chain rule, $$ \begin{aligned} \frac{d}{dx}\bigl(e^{bx}-1\bigr)&=b\,e^{bx}\,,\\ \frac{d}{dx}\ln(1+2x)&=\frac{2}{1+2x}\,. \end{aligned} $$

(a) The limit as an expression in $b.$ The last quotient is continuous at $0$ (Fact 5.17), so $$ \begin{aligned} &\lim_{x\to0}\frac{e^{bx}-1}{\ln(1+2x)}\\ &\quad=\lim_{x\to0}\frac{b\,e^{bx}}{\dfrac{2}{1+2x}}\\ &\quad=\lim_{x\to0}\frac{b\,e^{bx}(1+2x)}{2}\\ &\quad=\frac{b}{2}\,. \end{aligned} $$

(a) The parameter. The condition $\dfrac{b}{2}=3$ gives $$b=6\,.$$

(b) The conjugate. For $x>6$ the root is defined. Multiply and divide by the conjugate $x+\sqrt{x^{2}-6x}{:}$ $$ \begin{aligned} &x-\sqrt{x^{2}-6x}\\ &\quad=\frac{x^{2}-(x^{2}-6x)}{x+\sqrt{x^{2}-6x}}\\ &\quad=\frac{6x}{x+\sqrt{x^{2}-6x}}\,. \end{aligned} $$

(b) Divide by $x>0.$ As $x\to+\infty$ we may take $x>0,$ and then $x=\sqrt{x^{2}},$ so $\dfrac{\sqrt{x^{2}-6x}}{x}=\sqrt{1-\dfrac{6}{x}}.$ Dividing numerator and denominator by $x,$ $$ \begin{aligned} &\frac{6x}{x+\sqrt{x^{2}-6x}}\\ &\quad=\frac{6}{1+\sqrt{1-6/x}}\,. \end{aligned} $$

(b) Let $x\to+\infty.$ $\dfrac{6}{x}\to0,$ so the root tends to $1$ and the limit is $\dfrac{6}{1+1}{:}$ $$ \begin{aligned} &\lim_{x\to+\infty}\Bigl(x-\sqrt{x^{2}-6x}\Bigr)\\ &\quad=3\,. \end{aligned} $$

(a) Divide first. The numerator has higher degree than the denominator, so we divide with remainder (Fact 1.22). Polynomial division gives $$ \begin{aligned} &2x^{3}+x^{2}-4x+5\\ &\quad=(x+1)(2x^{2}-x-3)+8\,. \end{aligned} $$ Check by multiplying back: $$ \begin{aligned} &(x+1)(2x^{2}-x-3)\\ &\quad=2x^{3}+x^{2}-4x-3\,, \end{aligned} $$ and $-3+8=5.$

(a) A polynomial plus a remainder term. Hence $$ \begin{aligned} &\frac{2x^{3}+x^{2}-4x+5}{x+1}\\ &\quad=2x^{2}-x-3+\frac{8}{x+1}\,. \end{aligned} $$

(a) Integrate term by term. For $x>-1$ we have $x+1>0,$ so an antiderivative of $\dfrac{8}{x+1}$ is $8\ln(x+1)$ (no absolute value is needed). Hence, for $x>-1,$ $$ \begin{aligned} &\int\frac{2x^{3}+x^{2}-4x+5}{x+1}\,\mathrm{d}x\\ &\quad=\frac{2x^{3}}{3}-\frac{x^{2}}{2}-3x\\ &\qquad+8\ln(x+1)+C\,, \end{aligned} $$ with $C\in\mathbb{R}$ (Fact 7.13).

(b) The substitution. Put $u=-x^{3}+3x$ (Method 8.2); then $$ \begin{aligned} \mathrm{d}u&=(-3x^{2}+3)\,\mathrm{d}x\\ &=-3(x^{2}-1)\,\mathrm{d}x\,, \end{aligned} $$ so $(x^{2}-1)\,\mathrm{d}x=-\tfrac13\,\mathrm{d}u.$ The limits change with it: when $x=0,$ $u=0;$ when $x=s,$ $u=-s^{3}+3s.$

(b) On a finite interval $[0,s].$ $$ \begin{aligned} &\int_{0}^{s}(x^{2}-1)e^{-x^{3}+3x}\,\mathrm{d}x\\ &\quad=-\frac13\int_{0}^{-s^{3}+3s}e^{u}\,\mathrm{d}u\\ &\quad=-\frac13\Bigl[e^{u}\Bigr]_{0}^{-s^{3}+3s}\\ &\quad=\frac13\Bigl(1-e^{-s^{3}+3s}\Bigr)\,. \end{aligned} $$

(b) Let $s\to+\infty.$ The improper integral is the limit of this as $s\to+\infty$ (Definition 8.5). As $s\to+\infty,$ $-s^{3}+3s\to-\infty,$ so $e^{-s^{3}+3s}\to0.$ The improper integral converges and $$ \begin{aligned} &\int_{0}^{+\infty}(x^{2}-1)e^{-x^{3}+3x}\,\mathrm{d}x\\ &\quad=\frac13\,. \end{aligned} $$

(a) A cubic in $y.$ With $x=1$ the equation becomes $2y+y^{3}-1=2,$ that is $$y^{3}+2y-3=0\,.$$

(a) An integer root. An integer root divides $-3$ (Fact 1.18), so the candidates are $\pm1$ and $\pm3.$ Testing small integers, $y=1$ is a root: $1+2-3=0.$ By the factor theorem (Fact 1.17) $y-1$ is a factor, and $$ \begin{aligned} &y^{3}+2y-3\\ &\quad=(y-1)(y^{2}+y+3)\,. \end{aligned} $$

(a) The only value. The quadratic factor $y^{2}+y+3$ has discriminant $1-12=-11<0$ and therefore no real roots. So $$y(1)=1$$ is the only possible value.

(b) Differentiate the equation. Differentiating implicitly with respect to $x$ (Method 4.9) — the product rule on $2xy,$ the chain rule on $y^{3}$ — $$ \begin{aligned} &2y+2x\,y'+3y^{2}y'\\ &\quad-2x=0\,. \end{aligned} \tag{1} $$

(b) The slope and the tangent. At $x=1,$ $y=1{:}$ $2+2y'+3y'-2=0,$ so $5y'=0$ and $y'(1)=0.$ The tangent line at $(1,1)$ is $y=y(1)+y'(1)(x-1)$ (Fact 4.15), which is horizontal: $$y=1\,.$$

(c) The second derivative decides. Since $y'(1)=0,$ $x=1$ is a stationary point of $y(x),$ and the second derivative decides (Fact 7.3). Differentiating (1) once more (the product rule on $2x\,y'$ and on $3y^{2}\cdot y'$), $$ \begin{aligned} &2y'+2y'+2x\,y''\\ &\quad+6y\,(y')^{2}+3y^{2}y''\\ &\quad-2=0\,. \end{aligned} $$

(c) The verdict. At $x=1,$ $y=1,$ $y'=0{:}$ $2y''+3y''-2=0,$ so $$y''(1)=\frac25>0\,.$$ By the second-derivative test, $y(x)$ has a local minimum at $x=1$ (with value $y(1)=1$).

(a) Two conditions. $f'(x)=3x^{2}+2ax+b.$ A stationary point at $x=1$ means $f'(1)=0$ (Definition 6.12); with $f(1)=0$ the conditions are $$ \begin{aligned} 3+2a+b&=0\,,\\ 1+a+b+4&=0\,, \end{aligned} $$ that is $2a+b=-3$ and $a+b=-5.$

(a) Solve. Subtracting the second equation from the first, $a=2,$ and then $b=-5-a=-7{:}$ $$a=2\,,\quad b=-7\,.$$

(b) The derivative, factored. Now $f(x)=x^{3}+2x^{2}-7x+4$ and $$ \begin{aligned} f'(x)&=3x^{2}+4x-7\\ &=(3x+7)(x-1)\,, \end{aligned} $$ with zeros $x=-\tfrac73$ and $x=1.$

(b) Sign table (Method 2.25), one row per interval and per zero; ↗ means that $f$ increases, ↘ that it decreases:

(b) Increasing and decreasing. By Fact 2.23, $f$ is increasing on $\bigl(-\infty,-\tfrac73\bigr]$ and on $[1,+\infty),$ and decreasing on $\bigl[-\tfrac73,1\bigr].$

(b) The local extreme points. By the first-derivative test ($f'$ changes sign from $+$ to $-$ at $-\tfrac73$ and from $-$ to $+$ at $1$), $x=-\tfrac73$ is a local maximum point with $$f\bigl(-\tfrac73\bigr)=\frac{500}{27}\,,$$ and $x=1$ is a local minimum point with $f(1)=0.$ (Indeed $f(x)=(x-1)^{2}(x+4).$)

(c) The sign of $f''.$ $f''(x)=6x+4,$ which is negative for $x<-\tfrac23$ and positive for $x>-\tfrac23.$ Hence (Fact 6.20) $f$ is concave on $\bigl(-\infty,-\tfrac23\bigr]$ and convex on $\bigl[-\tfrac23,+\infty\bigr).$

(c) The inflection point. $f''$ changes sign at $x=-\tfrac23$ and nowhere else (Definition 7.6), and $f\bigl(-\tfrac23\bigr)=\tfrac{250}{27}.$ So the only inflection point is $$\Bigl(-\tfrac23,\ \tfrac{250}{27}\Bigr)\,.$$

(a) The first-order conditions (Fact 10.1): $$ \begin{aligned} f'_{x}&=2x+y-9=0\,,\\ f'_{y}&=x+2y-6=0\,. \end{aligned} $$ From the second, $x=6-2y;$ the first gives $12-4y+y-9=0,$ so $y=1$ and $x=4.$ The only stationary point is $(4,1),$ with $$ \begin{aligned} f(4,1)&=16+4+1-36-6\\ &=-21\,. \end{aligned} $$

(a) Classification. The second-order partial derivatives are constant: $A=f''_{xx}=2,$ $B=f''_{xy}=1,$ $C=f''_{yy}=2,$ so $$AC-B^{2}=3>0\,,\quad A>0\,,$$ and $(4,1)$ is a local minimum point (Fact 11.4).

(a) Global. Since $A>0,$ $C>0$ and $AC-B^{2}>0$ at every point, $f$ is convex on all of $\mathbb{R}^{2}$ (Fact 10.11), and a stationary point of a convex function is a global minimum point (Fact 10.13). So $f$ has the global minimum $$f(4,1)=-21\,.$$

(b) Existence, and one variable. $S$ is a segment: closed and bounded, and $f$ is continuous, so the extreme values exist (Fact 11.9) and are attained at a stationary point of the restriction in $(0,7)$ or at an endpoint. On $S,$ $y=7-x$ with $0\le x\le7$ (Method 10.19), and $$ \begin{aligned} g(x)&=f(x,7-x)\\ &=x^{2}+x(7-x)\\ &\quad+(7-x)^{2}-9x\\ &\quad-6(7-x)\\ &=x^{2}-10x+7\,. \end{aligned} $$

(b) Stationary point and endpoints (Method 6.15). $g'(x)=2x-10=0$ gives $x=5,$ inside $(0,7),$ and $$ \begin{aligned} g(5)&=-18\,,\quad g(0)=7\,,\\ g(7)&=-14\,. \end{aligned} $$ Therefore $\min_{S}f=-18$ at $(5,2)$ and $\max_{S}f=7$ at $(0,7).$

(c) The Lagrange conditions. With $\mathcal{L}=f(x,y)-\lambda(x+y-7)$ (Method 11.14), the first-order conditions are $$ \begin{aligned} 2x+y-9&=\lambda\,,\\ x+2y-6&=\lambda\,,\\ x+y&=7\,. \end{aligned} $$ Subtracting the first two gives $x-y-3=0,$ so $x=y+3;$ with $x+y=7$ we get $y=2,$ $x=5$ (the minimum point of (b)), and $$\lambda=2\cdot5+2-9=3\,.$$

(c) What $\lambda$ says. The multiplier is the rate of change of the optimal value with respect to the constraint level: if $f^{*}(c)$ denotes the minimum of $f$ on $x+y=c,$ then $\dfrac{df^{*}}{dc}=\lambda$ (Fact 11.21). Here: one more unit of $c$ raises the minimum value by about $3.$

(c) The estimate. With $dc=0.1,$ $$ \begin{aligned} f^{*}(7.1)&\approx f^{*}(7)+\lambda\cdot0.1\\ &=-18+0.3=-17.7\,. \end{aligned} $$ (The exact value is $-17.6925{:}$ with $x+y=7.1$ the conditions still give $x=y+3,$ so $(x,y)=(5.05,\,2.05),$ and the estimate is off by only $0.0075$ — for a small change the derivative is a good guide, Watch out 11.23.)

(a) The ratio and its domain. The series is geometric with ratio $k=\dfrac{\sqrt{x+2}}{3},$ which requires $x\ge-2,$ and then $k\ge0.$

(a) Convergence. It converges exactly when $|k|<1$ (Fact 8.7): $$ \begin{aligned} \frac{\sqrt{x+2}}{3}<1&\iff\sqrt{x+2}<3\\ &\iff x+2<9\\ &\iff x<7\,. \end{aligned} $$ So the series converges exactly for $$-2\le x<7\,.$$ (At $x=-2$ every term is $0;$ at $x=7$ every term is $1.$)

(a) The sum. The series starts at $n=1,$ so the first term is $k$ and the ratio $k;$ for such $x$ the sum is $\dfrac{k}{1-k}.$ Setting this equal to $\tfrac12{:}$ $2k=1-k,$ so $k=\tfrac13,$ that is $\sqrt{x+2}=1$ and $$x=-1\,,$$ which lies in the interval of convergence.

(b) The tail in closed form. The tail is a geometric series with first term $3^{-N}$ and ratio $\tfrac13,$ so $$ \begin{aligned} \sum_{k=N}^{\infty}3^{-k}&=\frac{3^{-N}}{1-\tfrac13}\\ &=\frac32\cdot3^{-N}\,. \end{aligned} $$

(b) The inequality. The condition $\tfrac32\cdot3^{-N}<3^{-2026}$ is equivalent to $$3^{2026-N}<\tfrac23\,.$$ Since $\tfrac23<1=3^{0}$ we need $2026-N<0,$ that is $N\ge2027;$ and $N=2027$ works because $3^{-1}=\tfrac13<\tfrac23,$ while $N=2026$ does not ($3^{0}=1>\tfrac23$). (Equivalently, taking logarithms: $N>2026+\ln(3/2)/\ln3\approx2026.37.$) So $$N=2027\,.$$

(a) Matrix form. With $$ \begin{aligned} \mathbf{A}&=\begin{pmatrix}2&1&-1\\1&3&2\\3&-1&1\end{pmatrix},\\[4pt] \mathbf{v}&=\begin{pmatrix}x\\y\\z\end{pmatrix},\quad \mathbf{b}=\begin{pmatrix}1\\13\\4\end{pmatrix}, \end{aligned} $$ the system is $\mathbf{A}\mathbf{v}=\mathbf{b}$ (Fact 12.21).

(a) The determinant. Expanding along the first row (Method 14.7), $$ \begin{aligned} \det\mathbf{A}&=2\begin{vmatrix}3&2\\-1&1\end{vmatrix} -1\begin{vmatrix}1&2\\3&1\end{vmatrix}\\ &\quad+(-1)\begin{vmatrix}1&3\\3&-1\end{vmatrix}\\ &=2\cdot5-(-5)-(-10)\\ &=25\,. \end{aligned} $$

(a) What it says. Since $\det\mathbf{A}=25\ne0,$ $\mathbf{A}$ is invertible and the system has exactly one solution (namely $\mathbf{v}=\mathbf{A}^{-1}\mathbf{b}$; Fact 14.18, Fact 14.26).

(b) Elimination (Method 13.16), writing $R_i$ for row $i.$ It is convenient to put the equation with leading coefficient $1$ first: $$ \begin{aligned} &\left(\begin{array}{ccc|c}2&1&-1&1\\1&3&2&13\\3&-1&1&4\end{array}\right)\\ &\quad\big\downarrow\ {\scriptstyle R_1\leftrightarrow R_2}\\ &\left(\begin{array}{ccc|c}1&3&2&13\\2&1&-1&1\\3&-1&1&4\end{array}\right)\\ &\quad\big\downarrow\ {\scriptstyle R_2-2R_1,\ R_3-3R_1}\\ &\left(\begin{array}{ccc|c}1&3&2&13\\0&-5&-5&-25\\0&-10&-5&-35\end{array}\right) \end{aligned} $$

(b) Elimination, continued. $$ \begin{aligned} &\quad\big\downarrow\ {\scriptstyle R_2/(-5),\ R_3/(-5)}\\ &\left(\begin{array}{ccc|c}1&3&2&13\\0&1&1&5\\0&2&1&7\end{array}\right)\\ &\quad\big\downarrow\ {\scriptstyle R_3-2R_2}\\ &\left(\begin{array}{ccc|c}1&3&2&13\\0&1&1&5\\0&0&-1&-3\end{array}\right). \end{aligned} $$

(b) Back substitution. The last row gives $-z=-3,$ so $z=3;$ then $y=5-z=2;$ and $x=13-3y-2z=1.$ So $$(x,y,z)=(1,\,2,\,3)\,.$$ Check: $2+2-3=1,$ $1+6+6=13,$ $3-2+3=4.$

(c) The criterion. A square matrix is not invertible exactly when its determinant is zero (Fact 14.18). Expanding along the first row, $$ \begin{aligned} &\begin{vmatrix}1&k&0\\k&4&1\\0&1&1\end{vmatrix}\\ &=1\cdot(4-1)-k\cdot(k-0)+0\\ &=3-k^{2}\,. \end{aligned} $$

(c) Both roots. This vanishes exactly when $k^{2}=3,$ so the matrix is not invertible for $$k=\sqrt3\ \text{ and }\ k=-\sqrt3\,.$$