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The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$, then some $c\in [a,b]$ satisfies $f(c)=y$.

Assume $f(a)<y<f(b)$ and letThe least-upper-bound property makes $c$ well defined. If $f(c)<y$, continuity gives points immediately to the right of $c$ in $S$, contradicting that $c$ is an upper bound. If $f(c)>y$, continuity gives a left neighborhood of $c$ disjoint from $S$, contradicting the definition of the supremum. Thus $f(c)=y$. The endpoint and reversed-order cases follow directly or by replacing $f$ with $-f$.

The mean value theorem states that if $g$ is continuous on $[\alpha ,\beta ]$ and differentiable on $(\alpha ,\beta )$, then some $c\in (\alpha ,\beta )$ satisfiesFor the functions in the question, the definition of the derivative makes $h$ continuous at $a$ and $f$ continuous at $b$, withIf the middle secant slope equals $k$, take $[\alpha ,\beta ]=[a,b]$. If it exceeds $k$, the intermediate value theorem applied to $h$ gives $\beta \in (a,b)$ with $h(\beta )=k$, and we take $\alpha =a$. If it is below $k$, apply the theorem to $f$ to obtain $\alpha \in (a,b)$ with $f(\alpha )=k$, and take $\beta =b$. In every caseThe mean value theorem on this subinterval then produces $c\in (\alpha ,\beta )\subset (a,b)$ withThis is the Darboux theorem for derivatives for the stated pair of derivative values.

First recall from the least-upper-bound property that every real interval is connected. For completeness, if $[a,b]=U\cup V$ were a separation with $a\in U$, let $s$ be the supremum of the points $t$ for which $[a,t]$ remains in $U$. Relative openness rules out both $s\in U$ and $s\in V$: in the first case one can move slightly to the right while staying in $U$, and in the second one finds points of $V$ slightly to the left of $s$, contrary to the definition of $s$. Thus no separation exists.

Part (a) now makes $g([0,1])$ a connected subset of the real line. Such a set is an interval: if it contained $u<v$ but omitted $y\in (u,v)$, its intersections with $(-\infty ,y)$ and $(y,\infty )$ would separate it. Since $g([0,1])$ contains $g(0)$ and $g(1)$, it consequently contains every number between them. Therefore for each such $y$ there is an $x\in [0,1]$ with $\boxed{g(x)=y}$. This derives the intermediate value theorem from connectedness.

For $p=0$ and $\lambda =1$, the linear ordinary differential equation is $y^{\prime \prime \prime \prime }=y$, whose general solution isBy the stated parity reduction, an even eigenfunction has the form $y=A\cos x+C\cosh x$. The conditions $y(c)=y^{\prime }(c)=0$ have a nonzero solution exactly whenthat is,An odd eigenfunction has the form $y=B\sin x+D\sinh x$, and the corresponding determinant givesThus $\lambda =1$ is an eigenvalue precisely under one of the two stated conditions.

For the minus sign, division by $\cos c\cosh c$ away from $c=\pi /2$ gives $\tan c=\tanh c$. On $(0,\pi /2)$, $\tan c>\tanh c$ for $c>0$, while on $(\pi /2,\pi )$ their signs differ, so there is no root. For the plus sign the equation isThere is no root in $(0,\pi /2)$. On $(\pi /2,\pi )$, the function $\tan c+\tanh c$ increases strictly from $-\infty$ to $\tanh \pi >0$, so it has exactly one root by the intermediate value theorem. Numerically,Hence the combined $\pm$ condition has exactly one solution in $0<c<\pi$.

In this discrete age-structured population model, individuals of age $a\ge 1$ present during the breeding season each produce $b_a$ newborns, so the number entering age zero isAn age-$a$ individual survives the following winter with probability $1-{\mu }_a$ and is then age $a+1$ in the next year, givingTogether with the initial age distribution, these equations determine every later cohort.

Substitute $n_{a,t}=r_a{\gamma }^t$ into the survival equation. Thenand iteration givesThe birth equation becomesA nonzero population mode therefore requires the Discrete Euler-Lotka equation

For nonnegative vital rates with at least one surviving reproductive age and enough decay to make the series finite, $\varphi$ is continuous and strictly decreasing on $\gamma >0$, tends to infinity as $\gamma \downarrow 0$, and tends to zero as $\gamma \to \infty$. The intermediate value theorem gives one positive solution, and strict monotonicity makes it unique.

The value $\varphi (1)$ is the expected lifetime offspring count under these fixed rates. Since $\varphi$ decreases, $\varphi (1)>1$ places the root at $\gamma >1$ and the population grows geometrically; $\varphi (1)<1$ places it at $\gamma <1$ and the population shrinks. Equality gives replacement, $\gamma =1$.

Differentiation givesThe initial conditions set the constant to one, sothe Pythagorean trigonometric identity.

Since $|c|\le 1$, Taylor theorem with Lagrange remainder givesfor some $0<\xi <1$, whilefor some $0<\eta <2$. The intermediate value theorem supplies $k\in (1,2)$ with $c(k)=0$. The sine addition formula gives $s(2k)=2s(k)c(k)=0$. The analogous cosine addition formula gives $c(2k)=c(k{)}^2-s(k{)}^2=-1$, soThus $s$ is a periodic function.

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$, then $f(c)=y$ for some $c\in [a,b]$.

Let $x_1<x_2<x_3$. If, for example, $f(x_2)$ exceeded both $f(x_1)$ and $f(x_3)$, choose $y$ strictly between $f(x_2)$ and $\max \{f(x_1),f(x_3)\}$. Applying the theorem on both $[x_1,x_2]$ and $[x_2,x_3]$ would give two distinct preimages of $y$, contradicting injectivity. The analogous argument excludes $f(x_2)$ below both endpoint values. Since the three values are distinct,

Fix $a<b$. If $f(a)<f(b)$, applying the displayed betweenness property to triples containing $a,b$ forces the same increasing order for every pair $x<y$; an order reversal would create a triple whose middle value is not between the other two. Thus $f$ is strictly increasing. If $f(a)>f(b)$, the same argument shows that it is strictly decreasing. This proves that every continuous bijection of the real line is monotone.

Since $\mathbb{C}$ is simply connected, writefor an entire function $f$. If $u$ were bounded above, thenwould be bounded, so the Liouville theorem would make $e^f$ constant. Its derivative $f^{\prime }{e}^f$ would then vanish, whence $f$ and $u$ would be constant. Applying the same argument to $e^{-f}$ rules out a lower bound. Thus a nonconstant $u$ is unbounded both above and below.

Given any $c\in \mathbb{R}$, choose $z_{-},z_{+}\in \mathbb{C}$ withThe restriction of $u$ to the line segment from $z_{-}$ to $z_{+}$ is continuous, so the intermediate value theorem supplies a point on that segment where $u=c$. ThereforeThis proves the surjectivity of a nonconstant entire harmonic function.

Order a period-three orbit as $a<b<c$. Its images produce two subintervals $I_0,I_1$ with covering relations $I_0\to I_0,I_1$ and $I_1\to I_0$. Any closed walk in this directed graph gives, by the interval-covering lemma and the intermediate value theorem, a fixed point of the corresponding iterate. Primitive closed walks of every length exist, giving an orbit of every positive period; this is the period-three case of Sharkovsky theorem.

Let $f:[a,b]\to \mathbb{R}$ be continuous. If it were unbounded, one could choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_j}\to x\in [a,b]$, but continuity would then give $f(x_{n_j})\to f(x)$, contradicting unboundedness.

Now let $M=\sup f([a,b])$. Choose $x_n$ with $f(x_n)>M-1/n$. A convergent subsequence and continuity give $f(x)=M$ at its limit. Applying the same argument to $-f$ shows that the infimum is attained. This proves the extreme value theorem on a closed bounded interval.

The functionis continuous and bounded, but attains neither its infimum $0$ nor its supremum $1$. For the second example, enumerate the rationals in $[0,1]$ as $(q_n)$ and setEvery nondegenerate interval contains infinitely many rational numbers, hence some $q_n$ with arbitrarily large $n$; therefore $\psi$ is unbounded on every such interval.

For the running extrema, compactness lets us writeThe function $f$ is uniformly continuous on $[a,b]$. Given $\epsilon >0$, choose $\delta >0$ such that $|f(u)-f(v)|<\epsilon$ whenever $|u-v|<\delta$. If $x<y$ and $y-x<\delta$, every new value $f(\xi )$ with $x\le \xi \le y$ is at most $f(x)+\epsilon \le M(x)+\epsilon$. Since $M(y)\ge M(x)$,Interchanging $x,y$ handles the other order. Applying this argument to $-f$ proves continuity of $m$ as well.

Finally fix $T>0$ and put $h(x)=g(x+T)-g(x)$. For each positive integer $n$, there must be some $x_n>n$ with $|h(x_n)|<1/n$. Otherwise, for some $n$ the continuous function $h$ would satisfy $|h(x)|\ge 1/n$ on the interval $(n,\infty )$. By the intermediate value theorem, $h$ would have a constant sign there. The valueswould then increase or decrease by at least $1/n$ at every step, contradicting boundedness of $g$. Thus $x_n\to \infty$ and

An interval map $F:I\to I$ has a horseshoe if there are two closed subintervals $J_0,J_1\subseteq I$ with disjoint interiors such thatIt is chaotic in Glendinning's sense if some positive iterate $F^n$ has a horseshoe.

Suppose $a<b<c$ are the points of a 3-cycle and put $J_0=[a,b]$, $J_1=[b,c]$. There are two possible cyclic orders. Ifthen the intermediate value theorem givesConsequently both $F^2(J_0)$ and $F^2(J_1)$ contain $J_0\cup J_1$. If insteadthenand again both second images contain $J_0\cup J_1$. Thus in either cyclic order $F^2$ has a horseshoe.

The closed interval does have the fixed-point property. If $f(x,0)=(F(x),0)$, then $F(0)-0\ge 0$ and $F(1)-1\le 0$. The intermediate value theorem supplies $x\in [0,1]$ with $F(x)-x=0$.

The intermediate value theorem says that a continuous $f:[a,b]\to \mathbb{R}$ takes every value between $f(a)$ and $f(b)$. For $f(a)<y<f(b)$, the closed sets where $f\le y$ and $f\ge y$ cannot separate the connected interval; equivalently, taking the supremum of $\{x:f(x)\le y\}$ and using continuity gives a point with value $y$.

Set $\varphi (a)=0$ and $\varphi (x)=\sin (1/(x-a))$ for $x>a$. Every interval $[a,b]$ contains a zero $c<b$; continuity on $[c,b]$ makes $\varphi$ take every value between $0=\varphi (a)$ and $\varphi (b)$, yet $\varphi$ is discontinuous at $a$.

A monotone function can be discontinuous only by a jump. If it had a jump at $c$, any number strictly between the left and right limits would lie between $f(a)$ and $f(b)$ but would not be attained, contrary to the hypothesis. Thus it is continuous.

For the last assertion, pass to subsequences with $x_n,y_n$ both within $1/n$ of $a$ and $g(x_n)$ near $l$, $g(y_n)$ near $L$. For $\lambda \in (l,L)$, the intermediate value theorem on the interval joining $x_n$ and $y_n$ gives $z_n$ with $g(z_n)=\lambda$; then $z_n\to a$. The endpoint cases use the original sequences.

Taking reciprocals in the finite product defining the gamma function givesLet $H_n={\sum }_{k=1}^n{k}^{-1}$. Sinceand $H_n-\log n\to \gamma$, where $\gamma$ is the Euler--Mascheroni constant, passage to the limit gives the Weierstrass product for the reciprocal gamma function

The logarithmic derivative of this identity isSince $k^{-1}-(z+k{)}^{-1}=z/[k(z+k)]$, the digamma function therefore satisfies

For real $z>0$, termwise differentiation gives the trigamma functionThus $\psi$ is strictly increasing on the positive real axis. At the two positive integers needed here, telescoping givesThe intermediate value theorem supplies a zero in $(1,2)$, and strict increase makes it unique. This is the positive zero of the digamma function.

For $0<\pi <1$, let ${\delta }_{\pi }$ be the Bayes classifier with class-zero prior $\pi$, and define its two class-conditional error probabilities byWriting the likelihood ratio as $L=f_1/f_0$, the rule chooses class one whenThe Gaussian likelihood-ratio level sets have probability zero, so dominated convergence shows that $e_0$ and $e_1$ depend continuously on $\pi$. As $\pi \downarrow 0$, the threshold tends to zero and the classifier chooses class one almost surely, givingAs $\pi \uparrow 1$, it chooses class zero almost surely, giving the opposite limit $(0,1)$. The intermediate value theorem therefore provides ${\pi }^{\ast }\in (0,1)$ such that

The rule ${\delta }_{{\pi }^{\ast }}$ is an equalizer rule with worst-case risk $r$. For any classifier $\delta$,because ${\delta }_{{\pi }^{\ast }}$ minimizes the integrated risk for its prior. Hence ${\delta }_{{\pi }^{\ast }}$ is a minimax decision rule, as summarized by the minimax Gaussian Bayes classifier from equal class errors.

The prior is indeed least favorable. Its Bayes risk is $r$, while for any other prior $q$ the Bayes risk is at most the integrated risk of the same equalizer rule ${\delta }_{{\pi }^{\ast }}$, namelyThus no prior has larger Bayes risk.

By the evenness of $V$, the semiclassical action integral is twice its integral over the positive half-axis. If $0\le E\le 1$, then

If $E\ge 1$, split the integral at $x=4$:The first integral isIn the second, put $r=\sqrt{x}$ and then $u=E+1-r$. It becomesAfter simplification,This is the action integral for a linear-to-square-root potential.

The graph begins at the origin, is continuous and strictly increasing, passes through $(1,16/3)$, and tends to infinity. Therefore for every $n\ge 0$ and $\epsilon >0$, the equationhas exactly one solution $E=E_n$ by the intermediate value theorem and strict monotonicity.

The Chebyshev alternation theorem says that a polynomial $P$ of degree at most $n$ is a best uniform approximation to $f\in C([0,1])$ if and only if there are $n+2$ pointsand a sign $\epsilon \in \{1,-1\}$ for whichThis is the equiripple criterion.

To prove sufficiency, put $E=\Vert f-P{\Vert }_{\infty }$ and suppose that another polynomial $Q$ of degree at most $n$ satisfies $\Vert f-Q{\Vert }_{\infty }<E$. At every $t_j$, the difference $R=Q-P$ must have the same sign as $f-P$; hence its signs alternate at the $n+2$ ordered points. The intermediate value theorem then gives at least one distinct root of $R$ in every interval $(t_j,t_{j+1})$, for at least $n+1$ roots in all. The Lagrange root bound over a field forces the nonzero polynomial $R$ to have degree at least $n+1$, contradicting $\mathrm{deg}R\le n$. Therefore no $Q$ gives a smaller error, and $P$ is best.

LetThe function $h$ is continuous, withThe intermediate value theorem gives $c\in [0,1]$ such that $h(c)=0$, and hence $f(c)=c$. This proves the fixed-point property of a closed interval.

Now suppose $0,1\in f([0,1])$. For any $y\in [0,1]$, choose $a,b\in [0,1]$ with $f(a)=0$ and $f(b)=1$. Applying the intermediate value theorem to $f$ on the interval with endpoints $a,b$ gives a point $x$ with $f(x)=y$. Thus $f$ is a surjective function.

Choose $z_{+},z_{-}\in \mathbb{C}$ withRestrict $f$ to the line segment between them:This is continuous, with $h(0)<0<h(1)$. The intermediate value theorem gives a $t_0$ with $h(t_0)=0$. Thereforeis the required zero on a line segment.

Define the continuous functionAt the mesh points its values telescope:If one term is zero, the result follows immediately. Otherwise the terms cannot all have the same sign, so there are mesh points at which $h$ is positive and negative. The intermediate value theorem between those points gives an $x\in [0,(n-1)a]$ with $h(x)=0$. Thus the telescoping increment lemma yields

The intermediate value theorem states that if $f:[\alpha ,\beta ]\to \mathbb{R}$ is continuous and $y$ lies between $f(\alpha )$ and $f(\beta )$, then some $c\in [\alpha ,\beta ]$ satisfies $f(c)=y$.

It is enough to prove the caseLetChoose $x_n\in E$ with $x_n\to c$. Continuity gives $f(c)\le y$. If $f(c)<y$, continuity would make $f(x)<y$ for some $x>c$, contradicting that $c$ is an upper bound of $E$. Thus $f(c)=y$. The case with reversed endpoint inequalities follows by replacing $f$ with $-f$.