By the even probability density of the standard normal distribution,The integral is finite. Equivalently, $|Z|$ has the half-normal distribution with unit scale.

For $n$ equal strips and a uniform bound $|f^{\prime \prime }(x)|\le M$ on $[a,b]$, the composite trapezoidal error bound isOn $[1,25]$, part (b) gives $M=1/2$: $f^{\prime \prime }$ is positive and decreases from $1/2$ after $x=1$. HenceTo certify error at most $10^{-4}$, choose $n\ge 2400$; equality gives exactly $10^{-4}$ in this bound. This is a sufficient certificate, not a claim that every smaller $n$ fails in actual numerical error.

Strictly, the maximum specified over the open interval $(1,25)$ is not attained; its supremum is $1/2$. Using the maximum on the closed interval, or simply the uniform bound $M=1/2$, gives the valid error estimate above.

With unit strip width, the composite trapezoidal rule givesFor Simpson's rule, the even integer arguments are the odd-numbered interior grid positions. ThereforeThe requested two-decimal answers are $T_{24}\approx 0.79$ and $S_{24}\approx 0.75$. The supplied component sums are individually rounded; they should not be treated as exact simultaneous identities. Computing from the original function gives approximately $0.786685$ and $0.748700$, respectively, with the same two-decimal results.

The printed inequality has its sign reversed: $f(x)<f_2(x)$ for $x>1$. This is present in the original scan, rather than arising in transcription. For $x>1$, Taylor's theorem with Lagrange remainder givesPart (a) has $f^{\prime \prime \prime }(\xi )<0$, so the remainder is strictly negative. As an independent check, direct algebra gives the quadratic Taylor majorant of a reciprocal quadratic:For example, $f(2)=1/5<1/4=f_2(2)$ is already a counterexample to the requested direction. The exact identity also shows that this polynomial majorizes $f$ on the entire real line, with equality only at one.

The Taylor polynomial uses $f(1)=1/2$, $f^{\prime }(1)=-1/2$ and $f^{\prime \prime }(1)=1/2$. ThereforeThe quadratic coefficient is $f^{\prime \prime }(1)/2!$, not $f^{\prime \prime }(1)$.

The critical points of $f^{\prime \prime }$ occur where $f^{\prime \prime \prime }=0$, namely $-1,0,1$. Its sign on the four intervals separated by these points is, in order, positive, negative, positive, negative. Thus $f^{\prime \prime }$ has local maxima at $\pm 1$ and a local minimum at zero. Their values areSince $f^{\prime \prime }(x)\to 0$ at both ends of the real line, these are also the global maximum $1/2$ and the global minimum $-2$. The sign analysis establishes extremality; merely solving $f^{\prime \prime \prime }=0$ would not.

Repeated use of the product rule and chain rule givesThe denominators are strictly positive on the whole real line, making the signs in the next parts easy to read.

The continuous function attains a global maximum and minimum on the closed disc by the extreme value theorem. All four unconstrained stationary points have squared distance eight, so none lies in its interior. The extrema must therefore lie on $x^2+y^2=6$.

The constraint gradient is nonzero on that circle, so Lagrange multipliers apply:Subtracting gives $(x-y)((x+y)/4-2\lambda )=0$. If $x=y$, the constraint gives $x=y=\pm \sqrt{3}$. Otherwise $2\lambda =(x+y)/4$; substitution into either equation gives $xy=-4$. This is impossible, since $|xy|\le (x^2+y^2)/2=3$ on the circle. Thus the diagonal points are the only candidates. Comparing their values gives

Write $f(x,y)=g(x)+g(y)$ with $g(s)=s^3/12-s$. On $[0,4]$, its derivative is $g^{\prime }(s)=s^2/4-1$, negative below two and positive above two. Compare the critical and endpoint values:Each coordinate can be chosen independently. Therefore the global minimum is $-8/3$ at $(2,2)$, and the global maximum is $8/3$ at $(4,4)$. This comparison includes the boundary of the rectangle, rather than classifying only interior stationary points.

Substituting $y=-x$ cancels both the cubic and linear terms, leaving $f(x,-x)=0$ for every real $x$. Thus the constrained maximum and minimum are both zero, attained everywhere on the constraint line. The feasible set is unbounded, but the restricted function is constant; compactness is not needed here.