Corrected practice final exam in differential calculus (201-SN2-RE)

Form $\frac{0}{0}$. Multiply by the conjugate: $\dfrac{\sqrt{x+1}-2}{x-3}\cdot\dfrac{\sqrt{x+1}+2}{\sqrt{x+1}+2}=\dfrac{(x+1)-4}{(x-3)(\sqrt{x+1}+2)}=\dfrac{1}{\sqrt{x+1}+2}$. At $x=3$: $\dfrac{1}{2+2}=\dfrac{1}{4}$.

Form $\infty-\infty$. The conjugate again: $\dfrac{(x^{2}+6x)-x^{2}}{\sqrt{x^{2}+6x}+x}=\dfrac{6x}{\sqrt{x^{2}+6x}+x}$. Divide the numerator and the denominator by $x>0$: $\dfrac{6}{\sqrt{1+6/x}+1}\longrightarrow\dfrac{6}{1+1}=3$.

Form $\frac{0}{0}$. Write $\dfrac{\sin 5x}{\tan 3x}=\dfrac{\sin 5x}{5x}\cdot\dfrac{3x}{\tan 3x}\cdot\dfrac{5}{3}$. The first two factors tend to 1, hence the limit $\dfrac{5}{3}$.

To the left of 2, $x-2<0$ so $|x-2|=2-x$. Then $\dfrac{(x-2)(x+2)}{2-x}=-(x+2)\longrightarrow-4$. To the right, $|x-2|=x-2$ and the expression equals $x+2\longrightarrow 4$. The two one-sided limits differ: the limit at 2 does not exist.

Form $1^{\infty}$. Let $y=\left(1+\frac{3}{x}\right)^{2x}$, so $\ln y=2x\ln\left(1+\frac{3}{x}\right)$. With $u=3/x\to 0$: $\ln y=\dfrac{6}{u}\ln(1+u)\to 6$ since $\frac{\ln(1+u)}{u}\to 1$. Therefore $y\to e^{6}\approx 403.43$.

From the right: $\lim_{x\to 1^{+}}f(x)=\ln 1+3=3$. From the left and at 1: $f(1)=a+b$. Continuity therefore requires $a+b=3$.

The one-sided derivatives: on the left, $f'(x)=2ax+b$ gives $2a+b$; on the right, $f'(x)=\frac{1}{x}$ gives 1. Differentiability requires $2a+b=1$. Subtracting the first equation: $a=1-3=-2$, then $b=3-a=5$. So $a=-2$, $b=5$, and $f'(1)=1$. (Note that differentiability only makes sense here once continuity is secured: without it, comparing the one-sided derivatives proves nothing.)

On the left, $f'(x)=-4x+5$ so $f''(x)=-4$: the left-hand limit of $f''$ is $-4$. On the right, $f'(x)=\frac{1}{x}$ so $f''(x)=-\frac{1}{x^{2}}$, which equals $-1$ at 1. The two values differ: $f'$ is not differentiable at 1, so $f$ is not twice differentiable at this point. On the other hand, $f'$ is defined and continuous there (it equals 1 on both sides): $f$ is of class $C^{1}$ but not $C^{2}$. The graph has a single tangent line at $(1,3)$, but its curvature jumps there.

$f$ is continuous on $[0,2]$ (a polynomial, then a logarithm, joined in question a) and $[0,2]$ is a closed, bounded interval: by the Extreme Value Theorem, $f$ reaches its maximum and its minimum there. On $[0,1]$, $f(x)=-2x^{2}+5x$ has its vertex at $x=\frac{5}{4}>1$, so $f$ is increasing there, from $f(0)=0$ to $f(1)=3$. On $(1,2]$, $\ln x+3$ is increasing, from 3 to $f(2)=3+\ln 2\approx 3.693$. So $f$ is increasing on the whole of $[0,2]$: minimum $f(0)=0$, maximum $f(2)=3+\ln 2\approx 3.69$.

$g$ is continuous on $[0,1]$ (product and sum of continuous functions). $g(0)=0\cdot e^{0}-2=-2<0$ and $g(1)=e-2\approx 0.718>0$. By the Intermediate Value Theorem, there exists $r\in(0,1)$ such that $g(r)=0$.

$g'(x)=e^{x}+x\,e^{x}=e^{x}(1+x)$, which is negative for $x<-1$: $g$ is therefore not monotonic on $\mathbb{R}$, and an argument based on monotonicity alone does not work. Split the line in two: for $x\leq 0$, $x\,e^{x}\leq 0<2$, so $g(x)<0$ and there is no root. For $x>0$, $g'(x)=e^{x}(1+x)>0$: $g$ is strictly increasing there, so it takes the value 0 at most once. Combined with a), the solution exists and is unique.

Newton: $x_{n+1}=x_{n}-\dfrac{x_{n}e^{x_{n}}-2}{e^{x_{n}}(1+x_{n})}$. From $x_{0}=1$: $g(1)=0.71828$, $g'(1)=2e=5.43656$, hence $x_{1}=1-\dfrac{0.71828}{5.43656}=0.86788$. Then $g(x_{1})=0.06716$, $g'(x_{1})=4.44902$, hence $x_{2}=0.86788-\dfrac{0.06716}{4.44902}=0.85278$.

$g(x_{2})=7.73\times 10^{-4}$ and $g'(x_{2})=e^{0.85278}(1.85278)=4.3470$. The estimate gives $|x_{2}-r|\approx\dfrac{7.73\times 10^{-4}}{4.3470}\approx 1.78\times 10^{-4}$. Three decimal places are therefore reliable: $r\approx 0.853$. (The exact value, $r=0.852606\ldots$, confirms an error of $1.78\times 10^{-4}$: the linear estimate is remarkably accurate here, because $g$ varies little over such a short interval.)

Product and chain rules: $f'(x)=5(3x^{2}-1)^{4}(6x)(2x+7)^{3}+(3x^{2}-1)^{5}\cdot 3(2x+7)^{2}(2)$. Factor out $(3x^{2}-1)^{4}(2x+7)^{2}$: $f'(x)=(3x^{2}-1)^{4}(2x+7)^{2}\left[30x(2x+7)+6(3x^{2}-1)\right]=(3x^{2}-1)^{4}(2x+7)^{2}(78x^{2}+210x-6)$, that is $f'(x)=6(3x^{2}-1)^{4}(2x+7)^{2}(13x^{2}+35x-1)$.

Chain rule with $\arctan'(u)=\frac{1}{1+u^{2}}$ and $u=\sqrt{x}$, $u'=\frac{1}{2\sqrt{x}}$: $g'(x)=\dfrac{1}{1+(\sqrt{x})^{2}}\cdot\dfrac{1}{2\sqrt{x}}=\dfrac{1}{2\sqrt{x}\,(1+x)}$.

Logarithmic differentiation. $\ln h=\sin x\cdot\ln x$, so $\dfrac{h'}{h}=\cos x\ln x+\dfrac{\sin x}{x}$, hence $h'(x)=x^{\sin x}\left(\cos x\,\ln x+\dfrac{\sin x}{x}\right)$. The power rule does not apply here: the exponent is not constant.

$y'=2e^{2x}\cos 3x-3e^{2x}\sin 3x=e^{2x}(2\cos 3x-3\sin 3x)$. Then $y''=2e^{2x}(2\cos 3x-3\sin 3x)+e^{2x}(-6\sin 3x-9\cos 3x)=e^{2x}(-5\cos 3x-12\sin 3x)$. Finally $y''-4y'+13y=e^{2x}\left[(-5-8+13)\cos 3x+(-12+12)\sin 3x\right]=0$.

Differentiate each term with respect to $x$, treating $y$ as a function of $x$: $2x-\left(y+x\dfrac{dy}{dx}\right)+2y\dfrac{dy}{dx}=0$. Collect the terms: $\dfrac{dy}{dx}(2y-x)=y-2x$, so $\dfrac{dy}{dx}=\dfrac{y-2x}{2y-x}$.

$1^{2}-(1)(3)+3^{2}=1-3+9=7$: the point is indeed on the curve. The slope is $\dfrac{3-2}{6-1}=\dfrac{1}{5}$, hence the tangent line $y-3=\dfrac{1}{5}(x-1)$, that is $y=\dfrac{x}{5}+\dfrac{14}{5}$.

Horizontal tangent: $\dfrac{dy}{dx}=0$, so $y=2x$ (with $2y-x\ne 0$, which holds since $2y-x=3x\ne 0$). Substitute into the equation: $x^{2}-2x^{2}+4x^{2}=7$, that is $3x^{2}=7$ and $x=\pm\sqrt{7/3}$. The points are $\left(\sqrt{\tfrac{7}{3}},\,2\sqrt{\tfrac{7}{3}}\right)\approx(1.53,\,3.06)$ and its reflection through the origin, $\left(-\sqrt{\tfrac{7}{3}},\,-2\sqrt{\tfrac{7}{3}}\right)$.

Swapping $x$ and $y$, the equation becomes $y^{2}-yx+x^{2}=7$, identical to the original: the curve is symmetric about the line $y=x$. This symmetry swaps the axes, so it turns every horizontal tangent line into a vertical one. The points we want are the mirror images of those in c), that is $\left(2\sqrt{\tfrac{7}{3}},\,\sqrt{\tfrac{7}{3}}\right)\approx(3.06,\,1.53)$ and its reflection through the origin. The formula confirms it: the denominator $2y-x$ is zero when $x=2y$.

$f'(x)=3x^{2}+3\geq 3>0$ for all $x$: $f$ is strictly increasing on $\mathbb{R}$, hence one-to-one. It is continuous (a polynomial), and $\lim_{x\to-\infty}f(x)=-\infty$, $\lim_{x\to+\infty}f(x)=+\infty$: by the Intermediate Value Theorem, every real number is reached, so $f$ is onto. It is indeed a bijection from $\mathbb{R}$ to $\mathbb{R}$.

We look for $a$ such that $f(a)=2$: $a^{3}+3a-2=2$, that is $a^{3}+3a-4=0$. The obvious root $a=1$ works ($1+3-4=0$), and it is the only one, since the function is strictly increasing. So $f^{-1}(2)=1$. Next, $f'(1)=3+3=6$, hence $\left(f^{-1}\right)'(2)=\dfrac{1}{6}$.

Let $u(x)=f^{-1}(x)$. The identity $f(u(x))=x$ holds on all of $\mathbb{R}$. Differentiate both sides with the chain rule: $f'(u(x))\cdot u'(x)=1$. Since $f'$ is never zero, isolate: $u'(x)=\dfrac{1}{f'(u(x))}$, that is $\left(f^{-1}\right)'(b)=\dfrac{1}{f'\left(f^{-1}(b)\right)}$.

Differentiate $f'(u)\,u'=1$ once more, with the product rule and the chain rule: $f''(u)\,(u')^{2}+f'(u)\,u''=0$, hence $u''=-\dfrac{f''(u)\,(u')^{2}}{f'(u)}=-\dfrac{f''(u)}{\left[f'(u)\right]^{3}}$ after replacing $u'$ by $1/f'(u)$. Here $f''(x)=6x$, so with $u(2)=1$: $\left(f^{-1}\right)''(2)=-\dfrac{f''(1)}{\left[f'(1)\right]^{3}}=-\dfrac{6}{6^{3}}=-\dfrac{1}{36}$. The negative sign says that $f^{-1}$ is concave down near 2, which is consistent: $f$ is concave up near 1.

Domain: $\mathbb{R}\setminus\{1\}$. Division: $x^{2}+3=(x+1)(x-1)+4$, so $f(x)=x+1+\dfrac{4}{x-1}$. Vertical asymptote $x=1$: as $x\to 1^{+}$, $\frac{4}{x-1}\to+\infty$ so $f\to+\infty$; as $x\to 1^{-}$, $f\to-\infty$. Oblique asymptote: $f(x)-(x+1)=\dfrac{4}{x-1}\to 0$ as $x\to\pm\infty$, so $y=x+1$ is an oblique asymptote at both ends. The curve lies above the asymptote for $x>1$ and below it for $x<1$.

$f'(x)=1-\dfrac{4}{(x-1)^{2}}=\dfrac{(x-1)^{2}-4}{(x-1)^{2}}=\dfrac{(x-3)(x+1)}{(x-1)^{2}}$. The denominator is always positive, so the sign is that of $(x-3)(x+1)$: $f'>0$ on $(-\infty,-1)$, $f'<0$ on $(-1,1)$ and $(1,3)$, $f'>0$ on $(3,+\infty)$. So $f$ increases, reaches a relative maximum at $x=-1$ with $f(-1)=\dfrac{4}{-2}=-2$, decreases down to the asymptote, comes back down from $+\infty$ after 1, reaches a relative minimum at $x=3$ with $f(3)=\dfrac{12}{2}=6$, then increases.

From $f'(x)=1-4(x-1)^{-2}$: $f''(x)=8(x-1)^{-3}=\dfrac{8}{(x-1)^{3}}$. It is never zero. For $x<1$, $f''<0$: the curve is concave down. For $x>1$, $f''>0$: concave up. The concavity does change at $x=1$, but $x=1$ is not in the domain: there is therefore no inflection point. This is the classic trap: a change of concavity is not enough, there must be a point on the curve.

There is no contradiction: the maximum and the minimum here are only relative, and they belong to two branches separated by the vertical asymptote. On the left branch, $f$ never goes above $-2$; on the right one, it never goes below 6. No theorem forces a local maximum on one branch to exceed a local minimum on another, precisely because the function is not continuous on $(-\infty,3]$. The sketch: two hyperbola-like branches on either side of $x=1$, each hugging $y=x+1$ at infinity, with the peak $(-1,-2)$ on the left and the trough $(3,6)$ on the right.

$f(x)=x^{1/3}$, $f'(x)=\frac{1}{3}x^{-2/3}$. At $a=125$: $f(125)=5$ and $f'(125)=\dfrac{1}{3\cdot 25}=\dfrac{1}{75}$. So $\sqrt[3]{124}\approx 5+\dfrac{124-125}{75}=5-\dfrac{1}{75}\approx 4.986667$. The calculator gives $4.986631$: the difference is $3.6\times 10^{-5}$, about seven millionths in relative terms. The linearization is excellent because $124$ is very close to $125$ and $f''$ is tiny there.

Take the logarithm before differentiating: $\ln T=\ln(2\pi)+\frac{1}{2}\ln L-\frac{1}{2}\ln g$. Differentiating, $\dfrac{dT}{T}=\dfrac{1}{2}\dfrac{dL}{L}$ (the term in $g$ disappears, $g$ being constant). In other words, a relative error on the length is only half passed on to the period: the square root damps the error.

$T=2\pi\sqrt{\dfrac{1.000}{9.8}}=2.0071$ s. Relative error: $\dfrac{dT}{T}=\dfrac{1}{2}\cdot\dfrac{0.005}{1.000}=0.0025=0.25\%$. Absolute error: $dT=0.0025\times 2.0071=0.0050$ s. (The exact computation gives $T(1.005)-T(1.000)=0.00501$ s: the differential is accurate to three digits.)