Not near the minimum at $1$. The derivative being set to zero factors asso $1$ has multiplicity of a root equal to three. The ordinary Newton iteration at a multiple root is generally only linearly convergent, with error ratio $(3-1)/3=2/3$ here.

In fact, cancellation in $x-f^{\prime }(x)/f^{\prime \prime }(x)$ gives the exact error identity, away from the original undefined quotient at the root,For iterates approaching $1$, the factor tends to $2/3$, proving linear convergence. The other zero of $f^{\prime }$ at $-1$ is simple and could instead give quadratic local convergence; the failure at $1$ is due to its multiplicity, not to every stationary point of this rational function.

Write $f^{(4)}(x)=24(5-4x)/x^6$. Its derivative is $f^{(5)}(x)=240(2x-3)/x^7$, so on $[1,2]$ it decreases to $x=3/2$ and then increases. Its endpoint and turning-point values areThus $M_4=24$, and the preceding bound becomesFor the rounding estimate, standard binary64 has spacing $ϵ=2^{-52}$ above one. Balancing this nominal absolute accuracy with the bound givesThus a few thousand even strips, roughly $5000$, reach the double-precision scale according to this bound. A scale-aware estimate leads to the same order. Indeed,lies in magnitude between $4$ and $8$, where the binary64 spacing is $2^{-50}$. Balancing the error bound against half that spacing gives about $4.16\times 10^3$ strips.

These are saturation estimates from balancing quadrature truncation and round-off error, not sharp guarantees of the actual rounding plateau. The error bound can be conservative, and evaluation and summation errors can limit accuracy earlier. Merely storing the final result does not specify every source of rounding error.

Let $h=(b-a)/n$ and let $M_4={\max }_{x\in [a,b]}|f^{(4)}(x)|$. A composite Simpson's rule panel covers two strips, so it has width $2h$ and absolute error at mostThere are $n/2$ panels. By the triangle inequality, the absolute error of their sum is at most the sum of their absolute errors. HenceThis assumes even $n$ and, for the usual Simpson remainder theorem, $f\in C^4[a,b]$. In particular its fourth derivative is bounded on the interval. It is the error bound for composite Simpson's rule, expressed in strip count rather than panel count.^[ref]

Substitution givesIn particular, $1$ is a stationary point. For $x$ sufficiently close to $1$, Taylor's theorem with fifth-order remainder giveswhere $\xi$ lies between $1$ and $x$. The fifth derivative is bounded near $1$, so the remainder is $O((x-1{)}^5)$ and is smaller in magnitude than half the positive quartic term for sufficiently small nonzero $x-1$. Therefore $x=1$ is a strict local minimum, despite the vanishing second derivative.

An exact algebraic check gives the stronger global conclusion:Equality holds only at $x=1$, so it is the unique global minimum on the actual domain.

The displayed rational expression has domain $\mathbb{R}\setminus \{0\}$; its value at zero is undefined despite the printed $\mathbb{R}\to \mathbb{R}$ notation. On that domain, the power rule givesThese derivatives are smooth on every interval avoiding zero, including all intervals used below.

Both terms $x^6$ and $\frac{3}{2}{y}^4$ are nonnegative, and they vanish simultaneously only at $(0,0)$. That point satisfies $x^4+2y^2\le 3$. Consequentlyand the global minimum is unique. No Lagrange multiplier calculation is needed: nonnegativity gives a lower bound attained at a feasible point.

Here the constraint surface $x^4+2y^2=3$ is compact and regular, so a global minimum exists and satisfies the Lagrange multiplier equationsKeep the zero-coordinate cases. If $y=0$, the two points are $(\pm 3^{1/4},0)$, with value $3\sqrt{3}$. If $x=0$, the two points are $(0,\pm \sqrt{3/2})$, with value $27/8$. If both coordinates are nonzero, the equations give $x^2=y^2=t>0$. The constraint becomes $t^2+2t=3$, whose only positive solution is $t=1$. These are the four points $(\pm 1,\pm 1)$, all with value $5/2$.

These cases exhaust the stated eight constrained stationary points. Comparing their values givesFor additional classification, substituting $u=x^2\in [0,\sqrt{3}]$ gives $F(u)=u^3+\frac{3}{8}(3-u^2{)}^2$, withIt decreases from $u=0$ to $u=1$ and increases from $u=1$ to $u=\sqrt{3}$. Thus the two $y$-axis points are constrained local maxima, the two $x$-axis points are global maxima, and the four mixed-coordinate points are global minima.

Use $F=x^6+\frac{3}{2}{y}^4$ and $G=x^4-2y^2-3$. Every feasible point has $x\ne 0$, so $\nabla G=(4x^3,-4y)$ cannot vanish. The Lagrange multiplier equations areThe first gives $\lambda =\frac{3}{2}{x}^2>0$. If $y\ne 0$, the second would give $\lambda =-\frac{3}{2}{y}^2<0$, a contradiction. Therefore $y=0$, and the constraint gives $x^4=3$. The only constrained stationary points are $(\pm 3^{1/4},0)$.

For a global certificate, feasibility implies $x^4\ge 3$, and henceEquality requires $y=0$ and $x^4=3$. ThusThe case split is essential: dividing by $y$ would discard precisely the minimizers.

Assume $f$ and $g$ are continuously differentiable near the point. At a regular feasible point, meaning $g(x)=0$ and $\nabla g(x)\ne 0$, a differentiable constrained local extremum satisfies the Lagrange multiplier equationsThey arise by differentiating the optimization Lagrangian $L(x,\lambda )=f(x)-\lambda g(x)$ with respect to all coordinates and the multiplier. Geometrically, the objective has zero derivative along the tangent space of the constraint surface, so its gradient must lie in the normal direction.

Solve these equations for candidates, examine feasible points where the regularity condition fails separately, and compare objective values or use a second-order test restricted to feasible directions. The equations locate stationary points; they do not by themselves establish which are minima or whether a global minimum is attained.

Let $e_n=x_n-x_{\ast }$ for the limit $x_{\ast }$. Quadratic convergence means that eventually $|e_{n+1}|\le C|e_n{|}^2$ for a fixed constant $C$; exact asymptotic order two has $|e_{n+1}|/|e_n{|}^2\to c$ with $0<c<\infty$.

For this iteration and $x_{\ast }=3$, direct algebra givesUsing the stipulated convergence $x_n\to 3$, the ratio of successive squared errors isThus the convergence is quadratic. The iteration is well defined here: $x_1=7/2$, and thereafter $x_n>3$ implies $3<x_{n+1}<x_n$. The initial point is not already the limit, so the error ratio is meaningful.