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A small difference between successive secant method iterates alone does not certify a small absolute error. Instead use a residual root error bound. On $[2,3]$,and $f(2)<0<f(3)$, so there is a unique root $r\in (2,3)$. For an iterate $x\in [2,3]$, the mean value theorem givesTherefore stop when $x\in [2,3]$ and $|x{\log }_{2}x-4|\le (1+1/\ln 2)10^{-10}$. This guarantees the desired absolute error in exact arithmetic; the evaluated residual of a root equation should include an allowance for its much smaller roundoff error. A safeguarded implementation can retain the bracket $[2,3]$ and use bisection if a proposed secant method step leaves it.

The second-order remainder form of Taylor's theorem gives, for a point ${\xi }_x$ between $0$ and $x$,For $x\ne 0$, the given curvature bounds implyIn particular $f(1)<-1/2<0$ and $f(2)\ge 5>0$. Differentiability implies continuity, so the intermediate value theorem gives a root in $(1,2)$.

The mean value theorem applied to $f^{\prime }$ gives $f^{\prime }(x)=3+x{f}^{\prime \prime }({\eta }_x)\ge 3+2x>0$ for $x\in (1,2)$. Thus $f$ is strictly increasing there, and it cannot have two roots. There is exactly one root in $(1,2)$.

Apply the mean value theorem, the first-order case of Taylor's theorem, to $f^{\prime }$ between $0$ and $x>1$. It yieldsAlso $f(3/2)\ge -5+9/2+9/4=7/4>0$, so the initial point is to the right of the root $x^{\ast }$. For an increasing strictly convex function, its tangent at a point to the right of a root meets the axis between that root and the point. Hence the exact Newton root-finding iteration remains in $[x^{\ast },3/2]$ and decreases towards $x^{\ast }$.

On every relevant interval, $|f^{\prime \prime }|<3$ and $f^{\prime }>5$, so the stated quadratic convergence bound gives, with $e_n=|x_n-x^{\ast }|$,Iterating the bound givesThus four Newton corrections suffice to reduce the exact-iteration error bound below the floating-point spacing scale. The quadratic convergence doubles the approximate number of correct digits at each late step. Beyond this point further corrections cannot reliably improve a generic floating-point answer. This is a bound on useful accuracy under the usual stable-evaluation model, not a claim that the data specify the exact iteration at which every implementation must stagnate.

For negative $x$, the mean value theorem gives $f^{\prime }(x)=3+x{f}^{\prime \prime }({\xi }_x)$. In particular,Since $f^{\prime }$ is continuous and strictly increasing, it vanishes at exactly one point $x_m\in (-2,-1)$. The positive second derivative makes $f$ a strictly convex function on the entire real line. Its derivative changes from negative to positive at $x_m$, so $x_m$ is the unique global minimum. One may also use $f(x)\ge x^2+3x-5\to \infty$ to verify coercivity.

In the endpoint-bracketing form of golden-section search, begin with $[-2,-1]$, a width-one interval known to contain this unique minimum, and place the two standard interior probes. The attainable position accuracy is different from that for a simple root. Around $x_m$, Taylor's theorem givesHere the minimum value is nonzero and of order one: the lower quadratic bound gives $f(x_m)\ge -29/4$, while $f(x_m)\le f(-1)<-13/2$. Ordinary relative round-off error in these function values therefore hides changes of order $ϵ$, once $h^2$ has that size. The function-value resolution near a nondegenerate minimum is consequently of orderEach golden-section search contraction multiplies the bracket width by $\rho =(\sqrt{5}-1)/2$. Using $2^{-20}$ as the nominal target width givesUnknown evaluation-error constants can change the actual stopping point; $29$ is the standard scale-based count. Requiring $2^{-40}$ position accuracy from these function-value comparisons would disregard the quadratic flatness of the minimum. The same precision effect is documented for one-dimensional numerical minimization.^[ref]

First control the denominator. The mean value theorem applied to $f^{\prime }$ yields, for $0<x\le 2000$,Thus $f$ is strictly increasing on $[0,2000]$ and all proposed Newton steps within that interval are defined.

Assume $0\le x_n<x^{\ast }$. Then $f(x_n)<0$ and $f^{\prime }(x_n)>0$, so $x_{n+1}>x_n$. Apply Taylor's theorem instead at the root, expanding about $x_n$:After rearranging and substituting the Newton step, the signed error satisfiesConsequently $x_n<x_{n+1}<x^{\ast }$. Starting from $x_0=0$, mathematical induction proves this for every finite $n$, establishing that the signed errors are negative and strictly increasing. Since $f^{\prime \prime }({\xi }_n)>-1/4000$ and $2f^{\prime }(x_n)>1$,Writing $E_n=-{ϵ}_n$ gives $0<E_{n+1}\le E_n^2/4000$. Since $E_0=x^{\ast }<2000$, this also implies $E_n\le 4000(E_0/4000{)}^{2^n}\to 0$, proving quadratic convergence. This is the concave Newton iteration from below.

Choose a radial mollifier $\rho \in C_c^{\infty }$ with integral one and support in the unit ball, and a radial cutoff function $\chi \in C_c^{\infty }$ equal to one there. For a radial tempered distribution $T$, the functionsare radial Schwartz functions. Smoothness comes from convolution of a tempered distribution with a Schwartz function, and compact support comes from the cutoff. Pairing with $\phi$ gives $⟨f_j,\phi ⟩=⟨T,{\check{\rho }}_{1/j}\ast (\chi (\cdot /j)\phi )⟩$. With $q_m(\phi )={\max }_{|\beta |\le m}{\sup }_x(1+|x|{)}^m|{\partial }^{\beta }\phi (x)|$, the cutoff tail and the mean value theorem giveThe continuity estimate for $T$ therefore proves $f_j\to T$ even in the strong dual topology. Inserting a cutoff is essential because convolution alone need not give rapid decay.

Suppose $f$ is continuous on an interval, differentiable in its interior, and $|f^{\prime }|\ge m>0$ throughout the interior. If $x_{\ast }$ is a root in the interval and $\hat{x}$ is another point in it, the mean value theorem gives $f(\hat{x})-f(x_{\ast })=f^{\prime }(\xi )(\hat{x}-x_{\ast })$ for an intermediate point $\xi$. Taking absolute values proves the stated bound from the residual of a root equation. This converts an observed residual into a stopping certificate, provided the evaluation error in that residual is also controlled.

Rolle's theorem is the equal-endpoint case of the mean value theorem. Let $a<b$ and let the real-valued function $f:[a,b]\to \mathbb{R}$ be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. If $f(a)=f(b)$, there exists $c\in (a,b)$ with $f^{\prime }(c)=0$. Geometrically, a tangent line somewhere in the interior is horizontal, like the secant line joining the endpoints. ^[ref]

For the polynomial $f(x)=x(x-2)$ on $[0,2]$, both endpoint values are zero and the derivative is $f^{\prime }(x)=2x-2$. Thus $c=1$, where $f(1)=-1$, and the horizontal tangent line is $y=-1$. The horizontal secant line is $y=0$: equal endpoint values do not force the interior tangent line to pass through those endpoints.

The absolute value function $f(x)=|x|$ on $[-1,1]$ exposes the importance of differentiability. It is continuous and $f(-1)=f(1)=1$, so the secant line has slope zero. However,and the derivative does not exist at zero because the one-sided derivatives disagree. Every existing derivative in the interior is nonzero. The corner fails a hypothesis of both Rolle's theorem and the mean value theorem, so neither conclusion is guaranteed.

For a short proof, the extreme value theorem supplies a greatest and a least value on $[a,b]$. If $f$ is a constant function, every interior point works. Otherwise at least one of these values differs from the common endpoint value, so it is attained at an interior extremum. At an interior local maximum or local minimum, differentiability forces the derivative to be zero: the difference quotients from the two sides have opposite weak signs and the same limit.

This turns repeated values into information about derivatives. Two distinct roots of a polynomial force a root of its derivative between them; iterating this idea gives the Rolle root count with multiplicities. More generally, a differentiable function with nonzero derivative throughout a real interval cannot repeat a value there, hence is injective. The beauty is that knowing only two equal values guarantees an exact local feature somewhere between them.

A twice-differentiable real-valued function on an open interval is convex if and only if its second derivative is nonnegative everywhere. If $f^{\prime \prime }\ge 0$, the mean value theorem makes $f^{\prime }$ nondecreasing, which implies the supporting-line and chord inequalities for convexity. Conversely, the slopes of a convex function are ordered; where the second derivative exists, this gives $f^{\prime \prime }\ge 0$. Strict positivity of $f^{\prime \prime }$ is sufficient for strict convexity, but is not necessary: $f(x)=x^4$ is strictly convex despite $f^{\prime \prime }(0)=0$.

A strictly distance-decreasing map decreases the distance between every pair of distinct points. Unlike a contraction mapping, it need not have a single Lipschitz constant below one. Two distinct fixed points would contradict strict decrease, so there is at most one. Existence can fail even on a complete metric space: on $[0,\infty )$, $T(x)=x+1/(1+x)$ has no fixed point, while the mean value theorem and $0\le T^{\prime }(x)<1$ show strict distance decrease.

For positive rates $a,b$, the mean value theorem in the rate gives a bound by $|a-b|x{e}^{-\min (a,b)x}$. Its maximum is $|a-b|/[e\min (a,b)]$. Combining this estimate with coefficient errors bounds a finite spectral sum uniformly on a half-line, including slow scales much longer than one. Positivity of both rates is essential: unstable exponentials need a different time-window or relative-error analysis.

If a twice differentiable real function on $\mathbb{R}$ satisfies $f^{\prime \prime }(x)\le -c$ for some fixed $c>0$, then it has a unique global maximum. The mean value theorem shows that $f^{\prime }$ is strictly decreasing, and that $f^{\prime }(x)+cx$ is nonincreasing. Hence $f^{\prime }(x)\le f^{\prime }(0)-cx$ for $x>0$ and $f^{\prime }(x)\ge f^{\prime }(0)-cx$ for $x<0$. These inequalities force opposite derivative signs at the two ends. The intermediate value theorem supplies a unique zero of $f^{\prime }$, at which $f$ changes from increasing to decreasing. Merely having $f^{\prime \prime }<0$ does not guarantee existence: $f(x)=-e^{-x}$ is increasing and approaches its unattained supremum zero.