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The binomial theorem gives coefficients $1,4,6,4,1$ for the fourth power. Substituting $2x$ for its variable gives

By the factor theorem, $x-2$ divides the polynomial. Division gives $2x^2+7x+3=(2x+1)(x+3)$, soIts three polynomial roots are $2,-1/2,-3$.

Writing the polynomial as $p(x)$, direct evaluation gives $p(0)=-6$ and $p(3)=42$. A polynomial is continuous, so the intermediate value theorem guarantees a polynomial root between them. Trying the integer $2$ gives $16+12-22-6=0$, hence $x=2$ is a root.

A tangent line through $(1,1)$ with slope $-2/3$ has equation

Implicit differentiation yields $3y^2{y}^{\prime }+2x=0$. At $x=1$, the real curve has $y^3=1$, hence $y=1$. Therefore

Differentiate each coordinate with respect to the parameter $t$ and use the chain rule:Both coordinates are defined where $\sin t\ne 0$ and $\cos t\ne 0$. As a check, $xy=1$ there, so the same derivative is $-1/x^2$.

Rewrite the exponential function as $2^x=e^{x\ln 2}$. The chain rule gives

The seven requested parabolas are $y=c{x}^2$ for $c=-3,-2,-1,0,1,2,3$; the case $c=0$ is the horizontal axis. Positive $c$ opens upward, negative $c$ downward, and larger $|c|$ gives a narrower parabola. All pass through the origin with a horizontal tangent line.

To find orthogonal trajectories, use the continuous family $y=c{x}^2$ containing these seven representative members. Its slope is $y^{\prime }=2cx=2y/x$ away from $x=0$. Two finite slopes of perpendicular tangents multiply to $-1$, so the orthogonal slope isThis separable differential equation gives $2ydy=-xdx$, henceThese ellipses are the regular real trajectories; $C=0$ is a single point, not a regular curve, and $C<0$ has no real points. One example is $x^2+2y^2=2$, with semiaxes $\sqrt{2}$ horizontally and $1$ vertically.

The implicit tangent relation $2xdx+4ydy=0$ also describes points with vertical tangents, where the slope quotient is unavailable. At intersections with the member $y=0$, the ellipse has a vertical tangent line and is perpendicular to that horizontal axis. The common origin of the parabolas is not on any regular trajectory with $C>0$. The sketch shows the requested seven members and the example ellipse.

The original differential equation is defined only where $x\ne 0$ and $y\ne 0$. Put $w=y^3$, so the chain rule gives $w^{\prime }=3y^2{y}^{\prime }$. Multiplying the equation by $3y^2$ yields a first-order linear ordinary differential equation:An integrating factor on either half-line is $1/x$, up to an irrelevant nonzero constant. Equivalently, apply the quotient rule directly:Integrating gives $w/x=x^2+C$. Thus the real general solution ison any connected interval where $x\ne 0$ and $x^3+Cx\ne 0$. These restrictions preserve both the original denominator and differentiability of the real cube root. In particular $C=0$ includes $y=x$ on either half-line; this solution has not been lost through division by an expression involving $y/x-1$.

For verification, differentiating $y^3=x^3+Cx$ gives $3y^2{y}^{\prime }=3x^2+C$, while $(2x^3+y^3)/x=3x^2+C$. The two sides agree wherever the original equation is defined.

Use integration by parts with $u=\ln x$ and $dv=xdx$. Since $du=dx/x$ and $v=x^2/2$,

The double-angle trigonometric identity $\sin 2x=2\sin x\cos x$ makes the integrand $2\sin x{\cos }^{2}x$. The integration by substitution $u=\cos x$, $du=-\sin xdx$, gives

The supplied bracket gives $0<e_0=1.5-\sqrt{2}<0.1$. Part (b) ensures $x_n>\sqrt{2}>1.4$, so each error satisfiesWriting $d_n=e_n/2.8$ gives $d_{n+1}<d_n^2$ and $d_0<0.1/2.8=1/28$. After three steps the exponent has doubled three times:This rigorous error bound uses only the given bracket, not advance knowledge of the digits of $\sqrt{2}$.

The actual rational iterates are $x_1=17/12$, $x_2=577/408$, and $x_3=665857/470832\approx 1.414213562374690$. They illustrate the bound, but the proof comes from the exact error identity and the positive lower bound on the denominators.

Write $r=\sqrt{a}>0$ and subtract it from the iteration:This is an Exact Newton error identity for a quadratic. If $x_n>r$, then $x_n>0$ and $e_n>0$, so $e_{n+1}>0$. Induction from $x_0>r$ proves $x_n>r$ for every $n$. Also $x_{n+1}-x_n=(a-x_n^2)/(2x_n)<0$, so the iterates decrease while remaining above the root. Their limit must satisfy $L=(L+a/L)/2$, hence $L=r$. The squared error numerator explains the ensuing quadratic convergence.

The Newton root-finding iteration follows the current tangent line to its horizontal-axis intersection. Since $f^{\prime }(x)=2x$, it givesA positive initial guess is appropriate for the positive square root; negative starting values instead target the negative root.

Using the signs chosen above, explicit multiplication givesThus the nonzero cases obey $A{\mathbf{v}}_m=\sqrt{{\sigma }_m}{\mathbf{u}}_m$. This relates the right singular vectors and left singular vectors in a singular value decomposition. For example, if $U=({\mathbf{u}}_1{\mathbf{u}}_2)$ and $V=({\mathbf{v}}_1{\mathbf{v}}_2{\mathbf{v}}_3)$, then $A=U\Sigma V^{\mathsf{T}}$ with $\Sigma =\left(\begin{array}{ccc}1& 0& 0\\ 0& \sqrt{2}& 0\end{array}\right)$.

In $A^{\mathsf{T}}A$, every row sums to one, so $(1,1,1)$ is an eigenvector with eigenvalue $1$. The antisymmetric direction $(1,0,-1)$ gives eigenvalue $2$. A direction orthogonal to both is $(1,-2,1)$, and direct multiplication shows that it lies in the kernel of a linear map. Normalizing all three gives the complete systemThey form an orthonormal basis of ${\mathbb{R}}^3$. The extra zero eigenvalue reflects the one-dimensional kernel of a linear map for this rank-two rectangular matrix. The question's symbols ${\sigma }_m$ denote the eigenvalues of $A^{\mathsf{T}}A$; the nonzero singular values of $A$ are their square roots, namely $1$ and $\sqrt{2}$.

For a matrix with equal diagonal entries, the symmetric and antisymmetric directions $(1,1)$ and $(1,-1)$ are eigenvectors. Direct multiplication of $A{A}^{\mathsf{T}}$ gives eigenvalues $3/2-1/2=1$ and $3/2+1/2=2$. Dividing each direction by its Euclidean norm givesThe two normalized eigenvectors are orthogonal; either may independently be replaced by its negative.

A row's squared Euclidean norm is $[(1+\sqrt{3}{)}^2+1+(1-\sqrt{3}{)}^2]/6=3/2$, and the dot product of the two rows is $[2(1-3)+1]/6=-1/2$. Thus the row Gram matrix isFor the column Gram matrix, the first and third squared column norms are $4/3$, the middle one is $1/3$, and the respective cross-products are $1/3,-2/3,1/3$. HenceThe two products are square matrices of different sizes; their unequal dimensions are consistent with $A$ being rectangular.