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The mean value theorem gives $f(b)-f(a)=(b-a)f\prime (c)>0$.

If $\alpha >1/2$, monotonicity already gives $\mathbb{E}{f}^{(1/2)}\ge \alpha >1/2$, so assume $\alpha \le 1/2$. Suppose for a contradiction that $\mathbb{E}{f}^{(1/2)}\le 1/2$. By the mean value theorem, some $s\in (p,1/2)$ satisfiesThe Margulis–Russo formula identifies this derivative with the appropriately normalized total influence, so ${\mathbf{I}}_s(f)$ is bounded solely in terms of $\zeta$. The $p$-biased Friedgut junta theorem then supplies, for any small $\eta >0$, a Boolean $J$-junta $h$ with $|J|\le r(\zeta ,\eta )$ and

Because $f$ is monotone, ${\mathbb{P}}_s(f=1)\le 1/2$, hence ${\mathbb{P}}_s(h=0)\ge 1/2-\eta \ge 1/4$ when $\eta \le 1/4$. It follows thatFor some assignment $u$ on $J$ with $h(u)=0$, therefore, ${\mathbb{E}}_s{f}_u\le 4\eta$. Monotonicity and $p<s$ imply ${\mathbb{E}}_p{f}_u\le 4\eta$. Choose $\eta <\alpha /5$, set $\epsilon =\eta$, and take $r\ge |J|$. Thencontradicting $(\epsilon ,p,r)$-quasirandomness. Thus $\mathbb{E}{f}^{(1/2)}>1/2$.

Choose $h_n=\min \{1/n,(b-c)/2\}>0$. Then $h_n\to 0$ and $[c,c+h_n]\subset (a,b)$. The mean value theorem states that a continuous function on a closed interval, differentiable in its interior, has an interior derivative equal to its endpoint difference quotient. It gives $x_n\in (c,c+h_n)$ withHere $x_n\ne c$, $x_n\to c$, and the right-hand side tends to $f^{\prime }(c)$ by differentiability at $c$. If $L={\lim }_{x\to c}{f}^{\prime }(x)$ exists, the same values $f^{\prime }(x_n)$ must tend to $L$. Uniqueness of the limit of a sequence givesThus a derivative cannot have a removable discontinuity at a point where the original function is differentiable. The proof also excludes an infinite limit of a function there, since the displayed difference quotients have a finite limit of a sequence.

The fixed-point iteration stays inside $[0,7]$ because $G$ maps that interval to itself. A fixed point satisfies $\xi =\sqrt{2+\xi }$, so $\xi \ge 0$ andOnly $\xi =2$ is admissible; $-1$ fails the unsquared equation. With $q=1/(2\sqrt{2})$, the mean value theorem gives the error boundInduction yields $|x_k-2|\le q^k|x_0-2|\le 5q^k\to 0$, since $0<q<1$. Therefore, for every allowed starting value,This estimate proves convergence directly, without assuming in advance that the sequence has a limit.

Set $H(x)=g(x)-x$. This function is continuous on $[a,b]$. Since $g$ takes values in $[a,b]$,The intermediate value theorem therefore supplies $\xi \in [a,b]$ with $H(\xi )=0$, equivalently $g(\xi )=\xi$. This includes the possibility that one endpoint is already a fixed point.

For uniqueness, the mean value theorem gives, for any $x,y\in [a,b]$,If both $\xi$ and $\eta$ are fixed points, thenSince $\gamma <1$, this forces $|\xi -\eta |=0$. Thus there exists exactly one fixed point. When $a=b$, the conclusion follows immediately from $g(a)=a$; otherwise the argument above applies. The contraction mapping estimate also proves convergence of the associated fixed-point iteration, as used next.

For $0\le x\le 7$, we have $2\le 2+x\le 9$. Taking the nonnegative square root givesThe derivative issoIn particular, the mean value theorem makes $G$ a contraction mapping on this interval.

The polynomial factors as $p(x)=(3x-2{)}^2(3x+1)$, so $\alpha =2/3$ is a multiple root of multiplicity two. The Newton root-finding iteration isFor $x\ne 0,2/3$, cancelling a factor givesThis expression extends continuously to $g(2/3)=2/3$. Let $I=[1/3,1]$. On this closed interval,Thus $g$ is increasing, andBy the mean value theorem, $|g(x)-g(y)|\le (16/27)|x-y|$ for all $x,y\in I$. The contraction mapping theorem states that a contraction mapping from a nonempty complete metric space into itself has a unique fixed point, and every fixed-point iteration converges to it. The closed interval $I$ with ordinary Euclidean distance is complete, and $g(2/3)=2/3$, soIn fact $|x_k-2/3|\le (16/27{)}^k|x_0-2/3|$. The Newton iteration at a multiple root here has linear convergence, since $g^{\prime }(2/3)=1/2$. Thus the convergence assertion is true for the simplified fixed-point iteration, and for the usual Newton root-finding iteration that stops upon reaching an exact root of a polynomial. If one insists on applying the unsimplified quotient at every step, it is undefined for the allowed starting value $x_0=2/3$, since both $p$ and $p^{\prime }$ vanish there; that literal version of an infinite sequence requires the stated stopping convention or continuous extension.

Take any $a<u<v<b$. By differentiability implies continuity, $f$ is continuous on $[u,v]$, and it is differentiable on $(u,v)$. The mean value theorem gives $c\in (u,v)$ for whichTherefore $\boxed{f(u)\le f(v)\text{ whenever }u<v}$: $f$ is a nondecreasing function, which is the non-strict meaning of increasing here.

The mean value theorem states: if a real-valued function $f$ is continuous on a closed interval $[u,v]$, where $u<v$, and differentiable on $(u,v)$, then there is $c\in (u,v)$ with

For any $u<v$ in the interval, the mean value theorem, applicable by differentiability implies continuity, givesAll pairs of values are equal. Thus $\boxed{f\text{ is a constant function}}$, the constancy theorem.

We prove the symmetric derivative criterion for monotonicity using an antiderivative. Fix $a<u<v<b$ and defineThe fundamental theorem of calculus gives $F^{\prime }=f$ because $f$ is continuous. For each interior $x$,Since $f^s(x)>0$, the definition of symmetric derivative gives $\delta >0$ such that $f(x+t)-f(x-t)>t{f}^s(x)$ for $0<t<\delta$. Therefore, for $0<h<\delta$,

We next show that $F$ is a convex function. On any closed interval $[r,s]\subset (a,b)$ let $\ell$ be the affine function joining $(r,F(r))$ to $(s,F(s))$. If $F-\ell$ were positive anywhere, the extreme value theorem would give a positive maximum at an interior point $x$. For every sufficiently small $h>0$, that maximum would implyThe affine function contributes zero to this symmetric difference, contradicting the strict positive inequality above. Thus $F\le \ell$ on every such interval, exactly convexity.

For a differentiable convex function, secant slopes bound endpoint derivatives. Explicitly, the chord inequality at $u+h$ and at $v-h$, followed by $h\downarrow 0$, givesSo $f$ is nondecreasing. If $f(u)=f(v)$ for some $u<v$, monotonicity would make $f$ constant on $[u,v]$, forcing $f^s(x)=0$ for $u<x<v$, a contradiction. Consequently $\boxed{f(u)<f(v)\text{ for every }u<v}$: $f$ is strictly increasing. This proof requires no continuity of $f^s$ and never applies the ordinary mean value theorem to a merely symmetrically differentiable function.

We verify that the augmented partition of an interval in part i has vanishing mesh of a partition. Define ${\varphi }_n(t)=t^2/(t^2+n)$ for $t\ge 0$. Its derivative isSince $2u\le 1+u^2$ for $u\ge 0$, we have $0\le {\varphi }_n^{\prime }(t)\le 1/\sqrt{n}$. The mean value theorem therefore givesThe appended interval has length $1/(n+1)$, so $\Vert {\mathcal{P}}_n\Vert \le \max \{n^{-1/2},(n+1{)}^{-1}\}\to 0$ for the augmented partition of an interval.

The sine is continuous, hence Riemann integrable on $[0,1]$. By the Riemann sum convergence theorem, $T_n\to {\int }_0^1\sin tdt=1-\cos 1$, using the fundamental theorem of calculus and the cosine antiderivative. As $T_n-S_n=\sin 1/(n+1)\to 0$, the answer is

Rearrange the polynomial equation as a fixed point equation:First establish existence and uniqueness of the target root of a polynomial. Since $f$ is a continuous function, $f(0)=-1<0$ and $f(1/7)=1/7^5>0$, the intermediate value theorem gives a root of a polynomial $\xi \in (0,1/7)$. Moreover $f^{\prime }(x)=5x^4+7>0$ for all real $x$, so $f$ is strictly increasing and this is its only real root of a polynomial.

For $x\in [0,1/7]$,so $g$ maps the closed interval into itself. AlsoThe mean value theorem therefore gives $|g(x)-g(y)|\le q|x-y|$ on the closed interval: $g$ is a contraction mapping. For any starting value $x_0\in [0,1/7]$, all iterates remain in that closed interval, and since $g(\xi )=\xi$,Thus the fixed-point iteration converges to the unique real root of a polynomial from every $x_0\in [0,1/7]$. This explicit error estimate proves the needed convergence directly.

The constancy theorem states that a differentiable function on an interval whose derivative vanishes everywhere is constant. At any included endpoints, continuity there is understood.

Indeed, take arbitrary $x<y$ in the interval. The mean value theorem applies on $[x,y]$ and gives $f(y)-f(x)=f^{\prime }(c)(y-x)=0$. Every pair of values is equal, proving the assertion. The interval hypothesis ensures that the whole segment $[x,y]$ lies in the domain.

The mean value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, where $a<b$, there is $c\in (a,b)$ such thatNo continuity assumption on the derivative is needed.

By the mean value theorem, $|f(x)-f(y)|\le A|x-y|$ for all real $x,y$. Thus $f$ is a contraction mapping on the complete metric space $\mathbb{R}$. Starting from arbitrary $x_1$, define $x_{n+1}=f(x_n)$. Induction gives $|x_{n+1}-x_n|\le A^{n-1}|x_2-x_1|$. For $m>n$, the triangle inequality and a geometric series yieldHence $(x_n)$ is a Cauchy sequence and completeness of the real numbers supplies a limit $x$. The Lipschitz continuity of $f$ implies $f(x)=\lim f(x_n)=\lim x_{n+1}=x$. Thus the limit is a fixed point, independently of the initial value. If $y$ were another, $|x-y|\le A|x-y|$ would force $x=y$. This proves the contraction mapping theorem in this setting, with error estimate

For the natural logarithm, $f^{\prime }(x)=1/x$. The chord slope is $(\log 2-\log 1)/(2-1)=\log 2$. Hence the mean value theorem point satisfies $1/\xi =\log 2$, givingIt lies strictly between one and two. For example, $\log 2={\int }_1^{2}du/u$ lies strictly between $1/2$ and one.

The chord through the endpoints is the straight line $y=(x-1)\log 2$. At $\xi =1/\log 2$, the tangent line isThese two straight lines have the same slope, as required by the mean value theorem. Since $f^{\prime \prime }(x)=-1/x^2<0$, the natural logarithm is a concave function: its graph lies above the chord and below this tangent line.

Assume $f\in C^{n+1}[a,b]$ and reorder the interpolation nodes increasingly as $x_0<\cdots <x_n$. The polynomial interpolation error $e=f-p_n$ vanishes at all of them. Rolle's theorem suppliesThese $n$ points are distinct because their containing intervals are disjoint. Thus $q=p_n^{\prime }$ is a polynomial of polynomial degree at most $n-1$ that interpolates the function $f^{\prime }$ at the interpolation nodes ${\eta }_1,\dots ,{\eta }_n$.

Apply the polynomial interpolation error theorem to $f^{\prime }$ with these $n$ interpolation nodes. Since the $n$th derivative of $f^{\prime }$ is $f^{(n+1)}$, for every $x\in [a,b]$ there is a ${\xi }_x\in (a,b)$ withFor $x={\eta }_i$ the equality is zero on both sides; otherwise the same auxiliary-function and repeated-Rolle's theorem proof applies. This also covers $n=1$, where $q$ is a constant polynomial and the proof is the ordinary mean value theorem for $f^{\prime }$.

Writing $M={\max }_{[a,b]}|f^{(n+1)}|$ gives the genuine error boundThe first formula is sharper, but its intermediate interpolation nodes depend on $f$; the last bound eliminates those unknown locations. This is derivative interpolation error from Rolle points.

For $0\le x\le 1$ and every real $z$,where $2|z|\le 1+z^2$ was used. The mean value theorem therefore proves Lipschitz continuity in $y$ with constant $1$.

Let $e_n=y(x_n)-y_n$. The explicit Euler method and the exact-solution identity from part (i) giveUsing the requested local truncation error bound $|{\tau }_n|\le h/2$, including at $n=0$, yieldsIteration and the finite geometric series giveSince $1+h\le e^h$ and $Nh=1$,Keeping the stronger bound $|{\tau }_n|\le h/4$ from part (i) improves the same proof to $(e-1)h/4$. The requested inequality follows already from the weaker bound.