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The following R function uses the factored polynomial for evaluation, handles endpoint and midpoint roots, and returns a midpoint only once its error bound is at most eps:

bisect <- function(a, b, eps = 1e-8, maxiter = 1000L) {
  f <- function(x) x * (x - 1) * (x - 2) * (x - 5)
  stopifnot(length(a) == 1L, length(b) == 1L,
            is.finite(a), is.finite(b), a < b,
            length(eps) == 1L, is.finite(eps), eps > 0)
  fa <- f(a)
  fb <- f(b)
  if (fa == 0) return(a)
  if (fb == 0) return(b)
  if (sign(fa) == sign(fb)) stop("Endpoints must bracket a sign change")
  for (i in seq_len(maxiter)) {
    m <- a + (b - a) / 2
    fm <- f(m)
    if (fm == 0 || (b - a) / 2 <= eps) return(m)
    if (sign(fm) == sign(fa)) {
      a <- m
      fa <- fm
    } else {
      b <- m
    }
  }
  stop("Iteration limit reached")
}

Each update preserves the sign-changing bracket, and the stopping bound is the half-width of that bracket. Thus continuity and the intermediate value theorem justify the returned root approximation.

Write $\rho =a\sqrt{2m{V}_0}/\hslash >0$. On the first tangent branch, $0<\beta <\pi /2$, the matching relations are equivalent toThe function $g(\beta )=\beta /\cos \beta$ is continuous and strictly increasing there:The intermediate value theorem therefore gives a unique first-branch root for every $\rho >0$. At this root $\beta <\rho$ and $\alpha =\beta \tan \beta >0$, so the energy eigenvalueobeys $-V_0<E<0$. This proves the even bound state in every attractive square well result.

To verify all the requested properties, take the resulting even wavefunction with exterior tails written in terms of their boundary values:The matching equation makes both the values and first derivatives agree at the endpoints. The exterior tails are square-integrable because $k>0$, and the interior interval is finite. More explicitly,so choosing $B$ as the positive inverse square root of the bracket gives wavefunction normalization. Thus an even, continuously differentiable, normalizable stationary solution always exists, however small the positive depth.

A fixed point $x_{\ast }$ satisfies $f(x_{\ast },\mu )=x_{\ast }$. A perturbation obeys $u_{n+1}=m{u}_n+O(u_n^2)$, where the periodic-orbit multiplier is $m=f_x(x_{\ast },\mu )$. It is linearly attracting for $|m|<1$ and repelling for $|m|>1$; at $m=\pm 1$ nonlinear terms decide Lyapunov stability. A crossing through $+1$ can produce a saddle-node, transcritical or pitchfork bifurcation, depending on the mathematical model parameter dependence and symmetry. A crossing through $-1$ generically gives a period-doubling bifurcation. A real one-dimensional function cannot have the complex conjugate pair needed for a Neimark–Sacker bifurcation.

For a genuine period-two orbit, study the return map $f\circ f$ at either point. By the chain rule its periodic-orbit multiplier isIts modulus determines linear stability by the same criterion, now per two iterates.

For the real quadratic iteration map, the fixed-point equation givesThe derivative is $-2x$. The lower branch $x_{-}$ is repelling whenever $\mu >-1/4$. The upper branch satisfies $|-2x_{+}|<1$ exactly forAt $\mu =-1/4$ the two fixed points merge in a saddle-node bifurcation. With $u=x+1/2$ at that mathematical model parameter, $u^{\prime }=u-u^2$, so the point attracts from the right and is repelling from the left. At $\mu =3/4$, $x_{+}=1/2$ has periodic-orbit multiplier $-1$ and undergoes a period-doubling bifurcation. At the threshold itself $u^{\prime }=-u-u^2$ and the two-step return is $u^{\prime \prime }=u-2u^3-u^4$, proving weak asymptotic stability there despite the inconclusive linear test.

Factor the second-iterate equation:Removing the fixed-point factor yields the periodic orbit pointsThey are distinct and exchanged by $f$ exactly when $\mu >3/4$; at equality they coincide with the fixed point. Their product is $1-\mu$, hence their periodic-orbit multiplier is $4(1-\mu )$. ThusIt is repelling for $\mu >5/4$ and reaches its next period-doubling bifurcation threshold at $\mu =5/4$. At that exact endpoint it is weakly attracting: if $f^2$ is expanded about either periodic orbit point as $v^{\prime }=-v+A{v}^2+B{v}^3+O(v^4)$, then $A=2\mu -6x_i^2$, $B=-4x_i$, and $A^2+B=20\pm 10\sqrt{2}>0$. Its second return is $v^{\prime \prime }=v-2(A^2+B)v^3+O(v^4)$.

At $\mu =2$ put $h(y)=4{\sin }^2(\pi y)-2=-2\cos (2\pi y)$. The double-angle identity givesThis verifies the printed relation to the Bernoulli shift. Precisely, it is a semiconjugacy: $h(y)=h(1-y)$, so it is not invertible on the circle. To obtain a genuine interval topological conjugacy, use the homeomorphism $H(u)=-2\cos (\pi u)$ from $[0,1]$ onto $[-2,2]$, and the full tent mapThe same identity proves $f(H(u),2)=H(T(u))$.

The full tent map has dense periodic points, is topologically mixing and has sensitive dependence on initial conditions. To see these properties directly, its $n$th iterate has $2^n$ monotone branches on dyadic intervals of length $2^{-n}$, each mapping onto $[0,1]$. Every nonempty open interval contains such a branch domain, and a later iterate covers all of $[0,1]$, proving topological mixing. On any branch, the function graph of the iterate has a fixed point by the intermediate value theorem, so periodic points occur in every open interval. Given any initial point and any neighborhood, some iterate of that neighborhood covers the interval; one can choose a nearby point whose image differs from the initial point's image by at least a fixed positive distance. This proves sensitive dependence on initial conditions. Uniform continuity of the inverse homeomorphism transfers these properties to $f$ on $[-2,2]$.

Consequently the real quadratic iteration map is chaotic on its invariant interval in the Devaney sense: topological transitivity, dense periodic points and sensitive dependence on initial conditions, with all these orbits bounded. There are also aperiodic orbits: periodic points form a countable union of finite sets of roots of an equation of the equations $f^n(x)=x$, whereas the interval is uncountable. The semiconjugacy alone would not justify treating $h$ as invertible; the tent-map argument supplies the missing rigorous transfer.

A homotopy from $f_0:Y\to X$ to $f_1:Y\to X$ is a continuous map $H:Y\times [0,1]\to X$ with $H(y,0)=f_0(y)$ and $H(y,1)=f_1(y)$ for every $y$. A covering map $p:\tilde{X}\to X$ is a surjective continuous map for which every $x\in X$ has an open neighborhood $U$ whose inverse image is a disjoint union of open sets, each mapped homeomorphically onto $U$. A lift of $f:Y\to X$ is a continuous map $\tilde{f}:Y\to \tilde{X}$ satisfying $p\circ \tilde{f}=f$.

The path lifting theorem says that a path $\gamma :[0,1]\to X$ and a point ${\tilde{x}}_0\in p^{-1}(\gamma (0))$ determine a unique path $\tilde{\gamma }:[0,1]\to \tilde{X}$ with $p\tilde{\gamma }=\gamma$ and $\tilde{\gamma }(0)={\tilde{x}}_0$.

On $S^1\setminus \{1\}$ there is a continuous branch ${\mathrm{Arg}}_{(0,2\pi )}$ of the complex argument, taking values in $(0,2\pi )$. If $f$ avoids $1$, set $\theta (z)={\mathrm{Arg}}_{(0,2\pi )}f(z)$. Then the explicit homotopystarts at $f$ and ends at the constant map $-1$. It is continuous because the chosen argument branch is continuous on the whole image. More generally, a map omitting any $w\in S^1$ is null-homotopic: apply the formula to $\overline{w}f$ and multiply the resulting homotopy by $w$.

For the based injective map, use the standard covering map $E(t)=e^{2\pi it}$. Lift the path $t\mapsto g(E(t))$ on $[0,1]$ to a real-valued path $G$ with $G(0)=0$. Since $g(1)=1$, its endpoint $d=G(1)$ is an integer. On $[0,1)$, $G$ is injective: equality $G(s)=G(t)$ implies $g(E(s))=g(E(t))$, then $E(s)=E(t)$ by injectivity, and hence $s=t$ in this half-open interval. A continuous injective real-valued function on an interval is strictly monotone.

If $G$ is increasing, $d>0$. It cannot be at least $2$, because the intermediate value theorem would produce $t\in (0,1)$ with $G(t)=1$, giving $g(E(t))=g(1)$ while $E(t)\ne 1$. Thus $d=1$. If $G$ is decreasing, the same reasoning gives $d=-1$. Now defineAt $t=0$ and $t=1$ this has the same value $1$ for every $s$, since $d$ is an integer. Therefore the formula descends continuously from $[0,1]\times [0,1]$ to $S^1\times [0,1]$. At $s=0$ it is $g$; at $s=1$ it is $z\mapsto z^d$. Hence $g$ is homotopic to the identity or to complex conjugation, with degree $1$ or $-1$. This is the key property of an injective self-map of a circle.

Finally, let $h$ be any continuous injective circle self-map. Normalize it to $g(z)=\overline{h(1)}h(z)$, which is injective and fixes $1$. If $h$ were not surjective, it would omit some circle point, so the explicit missing-point construction would make $h$ null-homotopic. Multiplying this homotopy by $\overline{h(1)}$ would make $g$ null-homotopic too. But $g$ is homotopic to the identity or conjugation, neither of which is null-homotopic by the permitted fact. This contradiction proves every continuous injective circle self-map is surjective. The lifting argument did not assume surjectivity, so the reasoning is not circular.

With local path connectedness, the intended conclusion holds. Choose a base point $x_0$ and the standard covering mapThe fundamental group of the circle is $\mathbb{Z}$. Since the image of a finite group under a group homomorphism is finite and $\mathbb{Z}$ has no nontrivial finite subgroups,is trivial. If $X$ is also locally path-connected, the lifting criterion for a covering space supplies a continuous map $\tilde{f}:X\to \mathbb{R}$ with $p\tilde{f}=f$, because $p_{\ast }{\pi }_1(\mathbb{R})=0$. The homotopyruns from $f$ to the constant function $f(x_0)$. This proves the claimed null-homotopy under that additional hypothesis, in particular for CW complexes and the topological graphs considered later in the paper.

For arbitrary path-connected spaces, the printed assertion is false. Here is an explicit Warsaw circle counterexample, including the obstruction that the usual lifting argument misses. In the Euclidean plane takeJoin $P=(1,0)$ to $Q=(0,-1)$ by the polygonal arc $A$ passing successively through $(2,0),(2,-2),(0,-2)$, and put $W=G\cup L\cup A$ with its subspace topology. The arc meets $G\cup L$ only at $P,Q$. The topologist's sine curve $G\cup L$ has two path components, but the added arc joins them, making $W$ path-connected.

Every loop in $W$ is contained infor some $\delta >0$. To justify the assertion, a path cannot reach $L$ along the oscillating tail: the intermediate value theorem for its $x$-coordinate would force its $y$-coordinate to attain both $1$ and $-1$ arbitrarily close to the supposed arrival time, contradicting continuity. The only alternative route from the tail to $L$ uses $P$ and the joining arc. If a loop visited tail points with $x\to 0$, a convergent subsequence of their parameter values would approach a time at which the loop lies in $L$. On a sufficiently small parameter interval around that time its image avoids $P$. That would give a forbidden path from $L$ into the tail without passing through $P$. Thus those tail coordinates have a positive lower bound. The space $T_{\delta }$ is a finite tree, in fact a single arc, so every such loop contracts. ConsequentlyNow define a continuous map $h:W\to S^1$ to be $1$ on $G\cup L$, and to make one full turn on $A$: for a parametrization $a:[0,1]\to A$ from $P$ to $Q$, set $h(a(t))=e^{2\pi it}$. The pieces $G\cup L$ and $A$ are closed in $W$, and the formulas agree at $P,Q$, so the pasting lemma for closed subspaces proves continuity.

There is no continuous lift $\tilde{h}:W\to \mathbb{R}$. Any such lift is integer-valued on $G\cup L$, so it is constant on each of the connected spaces $G,L$. Since $G$ accumulates on $L$, continuity forces those two constants to be the same. But along $A$ a lift of $e^{2\pi it}$ has the form $t+k$, giving values at $Q$ and $P$ differing by $1$, a contradiction. Finally, if $h$ had a null-homotopy, the homotopy lifting theorem for a covering map, starting with a lift of the final constant function and reversing time, would produce a continuous lift of $h$. This theorem needs no local path connectedness of the parameter space. Hence $h$ is not null-homotopic, although $W$ has finite, indeed trivial, fundamental group.

The native PDF explicitly assumes only path connectedness here. The missing local path connectedness cannot be inferred from the finiteness of the fundamental group.

For fixed $F>0$, the left side $L(v)=F^2{v}^2$ is a strictly increasing function on $v>1$, with limit $F^2$ at $1$ and unbounded growth. The right side $R(v)=1+3/[2(v-1)]+1/[2(v-1{)}^2]$ is a strictly decreasing function, sinceIt tends to infinity as $v\downarrow 1$ and to one as $v\to \infty$. Their continuity and the intermediate value theorem give a crossing, and strict opposite monotonicity makes it unique. There is exactly one admissible bore speed $V>U$ for every $F>0$.

The illustrated case is $F=1$; changing $F>0$ moves the increasing curve without changing the existence or uniqueness argument. The vertical asymptote at $v=1$ expresses the growing depth ratio $H/h_0=v/(v-1)$.

Since $x^2+1\ge 2|x|$, the image of a function is contained in $[-1/2,1/2]$. Both endpoints occur, at $x=-1$ and $x=1$. By continuity and the intermediate value theorem applied on $[-1,1]$, every intermediate value occurs too. HenceIt is not surjective onto $\mathbb{R}$ because, for example, $1$ is not attained. It is not injective because $f(2)=f(1/2)=2/5$ although $2\ne 1/2$.

Apply Newton's method to $F(t)=1-3\sin t/2-{\sin }^{3}t/3$. The derivative in part (a) givesChoose a starting point near a desired root, such as $t_0=0.6$ for the root in $(0,\pi /2)$; do not divide when $\cos t_k=0$. On $(0,\pi /2)$, $F$ is strictly decreasing and changes sign between its endpoints, so the intermediate value theorem gives exactly one root there. Its derivative is nonzero, giving the usual local convergence of Newton's method. The periodic function has other roots; the iteration does not impose global uniqueness or convergence from every starting value.

It is surjective but not injective; there is no global inverse function. Noninjectivity follows fromFor surjectivity, restrict to $x<1$, whereThis restriction is a continuous function, tends to $-\infty$ as $x\to -\infty$ and to $+\infty$ as $x\to 1^{-}$. The intermediate value theorem therefore supplies a preimage for every real number. The additional branch $x>1$ destroys global injectivity but cannot destroy surjectivity.

Not injective; surjective. Indeed, $f(0)=f(-1)=-9$ although $0\ne -1$. This polynomial is continuous, with $f(x)\to -\infty$ as $x\to -\infty$ and $f(x)\to \infty$ as $x\to \infty$. For any prescribed real $y$, choose $a<b$ with $f(a)<y<f(b)$; the intermediate value theorem supplies a preimage between $a$ and $b$.

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 $(1,2)$.

The coefficient assumptions imply a nonconstant polynomial with $p(0)<0$ and a positive leading coefficient. Thus $p(x)\to +\infty$ as $x\to +\infty$: dividing by its leading power of $x$ leaves the positive leading coefficient plus terms tending to zero. There is some $B>0$ with $p(B)>0$. A polynomial is continuous, so the intermediate value theorem supplies at least one positive root of a polynomial in $(0,B)$.

This gives neither uniqueness nor monotonicity. Multiple crossings are possible. If the degree of a polynomial is an even integer, the same leading-term argument gives $p(x)\to +\infty$ as $x\to -\infty$, so there is also a negative root of a polynomial. For an odd integer degree of a polynomial no negative root of a polynomial is forced; a linear polynomial with negative constant coefficient and positive slope is a counterexample.

Taylor's theorem at zero gives $f(x)=-1000+x+\frac{1}{2}{x}^2{f}^{\prime \prime }(\xi )$ for a point $\xi$ between zero and $x$. At $x=1000$, strict negativity of the second derivative gives $f(1000)<0$. At $x=2000$, its lower bound givesBy the intermediate value theorem there is a root strictly between these values. The given uniqueness identifies it as $1000<x^{\ast }<2000$.

A topological space $X$ is path-connected if, for every $a,b\in X$, there is a continuous mapThis continuous path must stay within $X$ for its entire duration. Its parameter need not represent physical time, and continuity does not require a derivative or finite arc length. ^[ref]

In the closed interval $[0,1]$, any $a,b$ are joined by $\gamma (t)=(1-t)a+tb$. For $a=1/4$ and $b=3/4$, this becomes $\gamma (t)=1/4+t/2$, so $\gamma (0)=1/4$, $\gamma (1/2)=1/2$, and $\gamma (1)=3/4$. All its values stay in $[0,1]$. In the union $D=[-2,-1]\cup [1,2]$, however, a continuous path from $-1$ to $1$ would attain $0$ by the intermediate value theorem, although $0\notin D$. Thus $D$ is not path-connected.

Every path-connected space is a connected space. Proof outline: if $X=U\cup V$ admitted a separation of a topological space, choose $a\in U$ and $b\in V$. A continuous path between them would have nonempty open inverse images ${\gamma }^{-1}(U),{\gamma }^{-1}(V)$ with empty intersection and union $[0,1]$. This contradicts the connectedness of that closed interval.

The converse fails: the topologist's sine curve is connected but has two path components. Path connectedness is useful when we need an actual continuous path of allowed positions, rather than only the absence of a separation of a topological space. For instance, when modelling allowed positions as a topological space, membership in the same path component answers whether one can move continuously between two positions while remaining in that topological space.

For a continuous function from an interval to itself with a three-cycle $a<b<c$, reflection if needed gives $F(a)=b,F(b)=c,F(c)=a$. Set $I=[a,b],J=[b,c]$. The intermediate value theorem gives $F(I)\supseteq J$ and $F(J)\supseteq I\cup J$. Successive closed-interval pullbacks realize every cyclic covering word by a point fixed under the corresponding iterate. The word $I{J}^{n-1}$ has a unique visit to the interior of $I$ per cycle, forcing least period $n$; the only boundary ambiguity is the original three-cycle, which realizes $n=3$ and cannot realize any other such word. A covering of $J$ by itself also gives a fixed point. Thus every positive least period occurs.

A wall moving at constant speed $U>0$ into stationary water of depth $h_0$ produces a hydraulic bore of speed $V>U$ in the uniform-state shallow water equations model. Behind the bore the fluid velocity equals $U$ and its depth $H$ obeys $H/h_0=V/(V-U)$. Put $F=U/\sqrt{g{h}_0}$ and $v=V/U$. The moving Rankine-Hugoniot conditions give $F^{2}v=[(H/h_0{)}^2-1]/2$, hence the displayed equation for $v>1$. Its left side is a strictly increasing function from $F^2$ to infinity, while its right side is a strictly decreasing function from infinity to one. The intermediate value theorem therefore gives exactly one admissible $V$ for every $F>0$.

For any nonzero positive drag, the cubic dust gravitational dispersion relation with gas drag is negative at $s=0$ if ${\omega }^2<{\Omega }^2$ and positive at sufficiently large positive $s$. The intermediate value theorem therefore guarantees a growing real root. This existence argument is independent of the weak-drag expansion and includes the dynamically unstable ${\omega }^2<0$ regime.

A real continuous function on $\mathbb{R}$ that tends to $+\infty$ as $|x|\to \infty$ has range $[m,\infty )$ for an attained minimum $m$. Choose a compact interval outside which its values exceed $f(0)$, and minimize there by the extreme value theorem. For any $y>m$, choose a far point with value larger than $y$ and apply the intermediate value theorem between it and a minimizer. Distance from a fixed point to a continuous graph is an example because it is at least the horizontal distance.

For rational offered load, every step of the stable recursion for the Erlang loss formula is rational. Comparing the final numerator and denominator with rational thresholds therefore certifies signs without floating-point uncertainty. Opposite certified signs of a continuous fixed-point residual imply a root by the intermediate value theorem; several alternating intervals can prove multiple Erlang fixed points under alternative routing.

A root bracket is a closed interval known to contain at least one zero of a function. For a real-valued continuous function on $[a,b]$, a strict change of sign between $f(a)$ and $f(b)$ certifies such an enclosure by the intermediate value theorem. An endpoint with value zero already supplies a zero of a function. For example, $[1,2]$ is a root bracket for the polynomial $f(x)=x^2-2$, since $f(1)=-1$ and $f(2)=2$. A root bracket need not contain a unique zero of a function.