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The proposed parametric equation traces precisely the same oriented segment, with a different speed. Set $u=t^n$, which increases from zero to one, and $du=n{t}^{n-1}dt$. Direct substitution gives the line integralsThus both values are unchanged. The derivative of $u$ vanishes at the initial endpoint for $n>1$, but the monotone integration by substitution is still valid. This is invariance under reparametrization of one curve, which applies even to a nonconservative vector field; it is not evidence of general path independence.

Apply the product rule to $x\sin (1/x)$ and the chain rule to the sine. Since $d(1/x)/dx=-1/x^2$, the derivative is

Use integration by substitution with $u=3-2x$, so $dx=-du/2$ and $x=(3-u)/2$. The indefinite integral becomesThus, with an arbitrary constant of integration,This real antiderivative applies for $x<3/2$ and extends continuously to the stated endpoint. Its left derivative there agrees with the limiting integrand, zero.

Use the derivative definition and the specified value $f(0)=0$:The last factor tends to one. Therefore the limit is zero when $p>0$, equals $A\omega$ when $p=0$, and diverges to $+\infty$ when $p<0$. A finite derivative exists exactly whenIn particular $-1<p<0$ gives continuity without differentiability. The threshold is an example of regularity of a power times a function with a simple zero.

Apply the chain rule twice: once to the square and once to the argument $x^3$ of the sine. The inner derivative is $y^{\prime }=3x^2\cos (x^3)-2\sin (2x)$. ThereforeThe original PDF has $\sin (x^3)$, rather than the cube of $\sin x$.

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 a right prism with equilateral triangular ends, the triangle area is $\sqrt{3}{\alpha }^2/4$. Thus its volume and external area areApply the Lagrange multiplier method to $V-\lambda (A-A_0)$ with positive lengths. The stationarity equations areThe second gives $\lambda =\sqrt{3}\alpha /12$. Substitution in the first then gives $6\beta =\sqrt{3}\alpha +3\beta$, or $\beta =\alpha /\sqrt{3}$. The area constraint becomesConsequentlyThe resulting volume is $V_{\max }=3^{3/4}/4{\mathrm{m}}^3$. As a check on the maximum, eliminating $\beta$ gives $V(\alpha )=\sqrt{3}{A}_0\alpha /12-{\alpha }^3/8$, whose second derivative is negative for every admissible positive $\alpha$; volume also tends to zero at both degenerate boundary shapes.

For the heat equation ${\Theta }_t=\kappa {\Theta }_{xx}$, separation of variables with homogeneous Neumann boundary condition at both ends gives spatial modes $\cos (n\pi x/L)$ and decay factors $e^{-\kappa (n\pi /L{)}^{2}t}$. The $n=0$ mode is constant and does not decay.

The initial profile already contains only two such modes:Evolving each mode therefore yieldsThe endpoint derivatives vanish, and substitution checks both the equation and the initial profile. The spatial integral remains $L{T}_0/2$: no heat escapes through insulated ends, so the limiting uniform temperature is $T_0/2$.

Substitute $r=-x^2$ into the geometric series, without taking any derivatives:Thus the coefficient of $x^n$ in the Taylor series is $(-1{)}^{n/2}$ when $n$ is even and zero when $n$ is odd. If nonzero terms are numbered from one, the $n$th term is $(-1{)}^{n-1}{x}^{2n-2}$.

For $a\ne 0$, factor out $|a|$, not $a$, and apply the binomial series with exponent $1/2$:For each fixed nonzero $a$, these are the first three nonzero terms of the Taylor series about zero; the full binomial series converges when $|x|<|a|$.

For $a=0$, the function is $|x|$. Its right derivative at zero is $1$ and its left derivative is $-1$, so no Taylor series about zero exists when $a=0$. The one-sided formulas are exactly $x$ for $x\ge 0$ and $-x$ for $x\le 0$; they do not form a single two-sided Taylor expansion, nor can they supply three nonzero terms. The parameter case explicitly included in the question must therefore be treated as this exception.

Here $t_0=0$ and $t_2=t_1^2$, so the constant and quadratic terms vanish, while the linear coefficient remains $t_1\ne 0$. A further derivative givesConsequently $f^{(3)}(0)=t_3-3t_1{t}_2+2t_1^3=t_3-t_1^3$. Since $t_3$ and $t_1$ have opposite signs, $t_3$ and $t_1^3$ also have opposite signs, and their difference is nonzero. The first two nonzero terms are thereforeThis is where the full sign condition $t_3{t}_1<0$ rules out a further cancellation in the Taylor expansion.

The difference and sum of the two antiderivatives in part (i) giveandTheir derivatives also verify the signs immediately.

The product rule givesFor $x>0$, the derivative is positive before $3$ and negative after $3$. Also $y(0)=0$ and $x^3{e}^{-x}\to 0$ as $x\to \infty$. Thus the global maximum on the requested range is

For $x>0$, the denominator is positive, so the function graph has the same sign as $\sin x$. Its crossings of the horizontal axis areEach crossing is genuine because the derivative there is $(-1{)}^n/(n\pi {)}^2\ne 0$. Near zero, $\sin x\sim x$, so $y\sim 1/x\to +\infty$; the origin is excluded and is not a crossing. The oscillations alternate sign and decay towards zero as $x\to \infty$, with absolute value at most $1/x^2$.

Here $C=0$ as well, but the branch through $5\pi /6$ is the other inverse-sine branch:Its derivative is $-1/\sqrt{1-x^2}$, and both the sign of $\tan y$ and the original equation agree with this choice.

The initial data force $C=0$, so the implicit relation factors as $\sin y(x-\sin y)=0$. Near $(0,0)$ this permits either the constant branch or the sine branch:Both satisfy the original equation, including its degenerate value $0=0$ at the initial point. They have different slopes there, zero and one, so the initial value problem is not unique. There are no further differentiable local branches through that point: away from $x=0$ the two factors cannot switch, and joining the two different branches at zero would give unequal one-sided derivatives.

The nonconstant branch is defined as a solution for $-1<x<1$; the constant branch exists for every real $x$.

On an interval with $x\ne 0$, division by $x^2$ gives the separable differential equation $y^{\prime }=(4/x^2-1/x)y^2$. For $y\ne 0$, take antiderivatives:Renaming the constant yields the explicit nonzero familyFor $x\ne 0$, the interval of definition must avoid zeros of the denominator. Division by $y^2$ excluded the additional solution $\boxed{y(x)\equiv 0}$, which must also be included.

The original undivided differential equation also makes sense at $x=0$, where it requires $y(0)=0$. For continuation through a zero of the leading ODE coefficient, rewrite the nonzero family asIt extends differentiably with $y(0)=0$ and $y^{\prime }(0)=1/4$, since $y(x)/x\to 1/4$. One may choose the integration constants independently on the two sides of zero and obtain a differentiable solution through zero. The zero solution also extends through zero, with derivative $0$; joining a nonzero branch on just one side to a zero branch on the other would fail differentiability.

The full cumulative distribution function isThus the function graph is horizontal at zero for $-1\le x\le 0$, rises continuously from $(0,0)$ to $(1,1)$, and is horizontal at one for $1\le x\le 2$. Its derivative $12x^2(1-x)$ is positive on $(0,1)$ and zero at both endpoints. The second derivative $12x(2-3x)$ changes sign at $x=2/3$, giving an inflection point there.