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The flat function $f$ is an even function, has its unique zero and global minimum at the origin, and approaches the horizontal asymptote $1$ as $x\to \pm \infty$. It decreases for negative $x$ and increases for positive $x$, meeting the origin with a horizontal tangent line.

Its derivative is an odd function: $f^{\prime }(x)=2x^{-3}{e}^{-1/x^2}$ off zero, with $f^{\prime }(0)=0$. It is negative to the left and positive to the right, tends to zero both at the origin and at infinity, and has one negative trough and one positive crest. Indeed $f^{\prime \prime }(x)=2e^{-1/x^2}(2-3x^2)/x^6$ away from zero, so the change in slope of $f^{\prime }$ occurs once on each side. The exact crest and trough coordinates are not needed for the sketch.

The curve has $x\ne 0$ and can be written $y=1/x-x$. It is a hyperbola with vertical asymptote $x=0$ and oblique asymptote $y=-x$, crossing the horizontal axis at $(\pm 1,0)$. It is symmetric under $(x,y)\mapsto (-x,-y)$.

On each branch $dy/dx=-1/x^2-1<0$. For $x>0$, the branch descends from $+\infty$ as $x\downarrow 0$, passes through $(1,0)$, and approaches $y=-x$ from above as $x\to +\infty$. The negative branch follows by the central symmetry.

The domain of a function is $\mathbb{R}\setminus \{0\}$, because the denominator vanishes at zero. Near zero, $e^x-1\sim x$, so $x=0$ is a vertical asymptote, approached with value $-\infty$ from the left and $+\infty$ from the right. The horizontal asymptotes are $y=-1$ as $x\to -\infty$ and $y=0$ as $x\to +\infty$.

Moreover,Both branches define a strictly decreasing monotone function: the left branch lies below $-1$, and the right branch is positive. There are no zeros or vertical-axis intercept.

The tangent has poles at $\pi /2$ and $3\pi /2$, so these are vertical asymptotes of $x\tan x$. The zeros on the given interval are $0,\pi ,2\pi$. Near zero, $x\tan x=x^2+O(x^4)$, so the graph starts with a horizontal tangent line.

Each of its three branches is increasing. Indeedand the numerator starts at zero with derivative $2{\cos }^{2}x\ge 0$, so it is positive for $x>0$. The first branch rises from zero to $+\infty$, the middle branch rises from $-\infty$ through $(\pi ,0)$ to $+\infty$, and the last rises from $-\infty$ to $(2\pi ,0)$.

Solving for $y$ gives $y=\pm (b/a)\sqrt{x^2-a^2}$ with $x\ge a$. For large $x$, the square root is $x-a^2/(2x)+O(x^{-3})$, so the two asymptotes areThe curve lies between these lines on the right side and has its nearest point, with vertical tangent, at $(a,0)$. As $t$ increases it travels from the lower unbounded end through that vertex to the upper unbounded end. The left hyperbola branch is not part of the given curve.

Where the original expression is defined, multiplication by $\cos x$ givesThus the graph repeats with period $\pi$ and has vertical asymptotesBetween consecutive asymptotes the extended cotangent branch decreases from $+\infty$ to $-\infty$, since $y^{\prime }=-1/(\cos x+\sin x{)}^2<0$. It passes through $(k\pi ,1)$ and $(\pi /4+k\pi ,1/2)$.

It is essential to preserve the domain when simplifying a function. At $x=\pi /2+k\pi$, the original tangent is undefined. The finite limiting value is zero, so the graph has holes at $(\pi /2+k\pi ,0)$, rather than actual zeros. The simplified expression extends across those points, but the original one does not.

This exponential function is positive everywhere, with $y(0)=1$, $y\to \infty$ as $x\to -\infty$ and $y\to 0$ as $x\to \infty$. Its derivatives areIt is strictly decreasing overall, with a horizontal stationary inflection point at $(0,1)$. A second, nonstationary inflection point occurs at $x=(2/3{)}^{1/3}$, where $y=e^{-2/3}$. The only horizontal asymptote is $y=0$ toward positive infinity; the graph is not symmetric about the vertical axis.

Within the requested interval, the function is symmetric about $x=\pi /2$ because $\sin (\pi -x)=\sin x$. Its derivative $\cot x$ is positive before $\pi /2$, zero there, and negative afterwards. The negative second derivative shows that this is a strictly concave function. Consequently its unique maximum is $(\pi /2,0)$. It tends to $-\infty$ at both ends, giving vertical asymptotes $x=0$ and $x=\pi$.

Rewrite the rational function as $y=1-1/(x^3+1)$. It is undefined at $x=-1$, with the vertical asymptote $x=-1$, and tends to the horizontal asymptote $y=1$ as $x\to \pm \infty$. At the pole, $y\to +\infty$ from the left and $y\to -\infty$ from the right.

Its derivative isvanishing only at $x=0$. Since $y(0)=0$ and $y=x^3+O(x^6)$ near zero, this is a stationary inflection point, not an extremum. ThusThe graph increases on each side of the vertical asymptote. The left branch lies above $y=1$, while the right branch lies below it and crosses the origin horizontally.

Take $a>0$ for the sketch; changing its sign rotates the picture by $\pi$. The parametric curve passes through $(0,0)$ at $t=0$ and through $(3a/2,3a/2)$ at $t=1$. As $t\to \pm \infty$ it returns towards the origin, since $x\sim 3a/t^2$ and $y\sim 3a/t$. The positive-$t$ branch makes a loop in the first quadrant, leaving the origin tangent to the positive $x$ axis and returning tangent to the positive $y$ axis.

For a more precise sketch, differentiation givesThe loop has its largest $x$ at $t=2^{-1/3}$ and largest $y$ at $t=2^{1/3}$. Their coordinates are $a(2^{2/3},2^{1/3})$ and its reflection in $y=x$.

The parameter value $t=-1$ is excluded. Cancelling the common factor in the sum givesThus the asymptote is $\boxed{x+y+a=0}$. As $t\to -1^{-}$, $x\to +\infty$ and $y\to -\infty$; as $t\to -1^{+}$, $x\to -\infty$ and $y\to +\infty$. The branch $t<-1$ approaches the origin from the fourth quadrant as $t\to -\infty$, while $-1<t<0$ connects the other infinite branch to the origin through the second quadrant.

This is the folium of Descartes. If $a=0$, the parametrization degenerates to the origin for every defined parameter value and there is no nontrivial loop.

Differentiate the two powers to locate the stationary point:For $x>0$, this is negative before $2$ and positive afterwards, so the global minimum isThe curve crosses the horizontal axis at $x=1$. Its limits are $+\infty$ as $x\to 0^{+}$ and $0$ from below as $x\to \infty$, giving the vertical asymptote $x=0$ and horizontal asymptote $y=0$. These features determine the required sketch.

The graph is the upper branch of the hyperbola $d^2-(t+1{)}^2=1/2$. It is symmetric about $t=-1$, positive everywhere, and has its lowest point at $(-1,1/\sqrt{2})$. Its derivatives areso it is strictly convex. Sinceits two oblique asymptotes are

For the requested parameter value $a=2$, the function graph has two branches, symmetric about the vertical axis because $f$ is an even function. Its derivatives areThe stationary points are $\boxed{(-2,2)\text{ and }(2,2)}$, both local minima and also the lowest points of their respective branches. The function tends to $+\infty$ as $x\to 0^{\pm }$, so $x=0$ is a vertical asymptote; it also tends to $+\infty$ as $|x|\to \infty$. The negative branch decreases to $(-2,2)$ and then rises towards the vertical asymptote; the positive branch falls from the vertical asymptote to $(2,2)$ and then rises.

Introduce the dimensionless quantities $\tau =\mu t/r_0$ and $w=\mu v/(g{r}_0)$. ThenThe Taylor series at $t=0$ givesThus the initial slope is $g$, as for free fall. At late times,The oblique asymptote is $v=gt/4+g{r}_0/(4\mu )$, approached from below. Direct differentiation givesThe speed increases monotonically with decreasing slope, from $g$ towards $g/4$. It does not reach a terminal velocity in this drag-free model. Without mass accumulation, and still without drag, a drop released from rest has $v=gt$: its acceleration remains $g$, whereas the growing drop continually shares its momentum with newly acquired material.

Assume a real, time-independent potential with the same zero asymptote on both sides, so $k=\sqrt{2mE}/\hslash$ is the same incident and transmitted wave number. For unit incoming amplitudes,Multiplying the two Schrödinger equations crosswise and subtracting shows that the Wronskian $W={\psi }_L{\psi }_R^{\prime }-{\psi }_L^{\prime }{\psi }_R$ is constant. Its two asymptotic values are $-2ik{t}_R$ and $-2ik{t}_L$, hence $t_L=t_R$.

For a real potential the probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ is conserved, giving $|r_L{|}^2+|t_L{|}^2=|r_R{|}^2+|t_R{|}^2=1$. ThereforeThe reflection amplitudes can have different phases. Reality or self-adjointness is essential for the reflection-probability conclusion: a complex absorbing potential need not have equal left and right reflectances. Equal asymptotic speeds are also used in identifying transmission with the squared amplitude.