Proof.

1.

Suppose \(f\) is not bounded. Then for every \(n\in \mathbb {N}\) there is an \(x_n \in [a,b]\) such that \begin {equation} \label {eq:unbdd} |f(x_n)| > n . \end {equation} The sequence \(\{x_n\}\) lies in \([a,b]\) and so is bounded. By the Bolzano–Weierstrass theorem (for sequences) it has a convergent subsequence \(\{x_{n_k}\}\), with limit \(x\) say. Since \([a,b]\) is closed, \(x\in [a,b]\). Putting \(n = n_k\) in (2.7) gives \(|f(x_{n_k})| > n_k \geq k\) for every \(k\), so \(\{f(x_{n_k})\}\) is unbounded and hence not convergent. But \(f\) is continuous at \(x\), so \(f(x_{n_k}) \longrightarrow f(x)\), a contradiction. Hence \(f\) is bounded on \([a,b]\).

2.

By part 1, \(M = \sup \{f(x): x\in [a,b]\}\) exists. For each \(n\in \mathbb {N}\), \(M - \frac {1}{n}\) is not an upper bound, so there is an \(x_n \in [a,b]\) with \[M - \frac {1}{n} < f(x_n) \leq M,\] and therefore \begin {equation} \label {eq:fxnM} \lim _{n\rightarrow \infty } f(x_n) = M . \end {equation} The sequence \(\{x_n\}\) is bounded, so by the Bolzano–Weierstrass theorem it has a convergent subsequence \(\{x_{n_k}\}\), with limit \(\alpha \) say. Since \([a,b]\) is closed, \(\alpha \in [a,b]\), and since \(f\) is continuous at \(\alpha \), \(f( x_{n_k})\longrightarrow f(\alpha )\). On the other hand, by (2.8), \(f(x_{n_k})\longrightarrow M\). A sequence cannot converge to two different limits, so \(f(\alpha ) = M\). The proof that \(f(\beta ) = \inf \{ f(x): x \in [a,b]\}\) for some \(\beta \) is similar.

Proof. Take \(\varepsilon = |f(p)|>0\) in the definition of continuity: there is a \(\delta > 0\) such that \[f(p) - |f(p)| < f(x) < f(p) + |f(p)| \qquad \text {whenever } x\in [a,b] \text { and } |x - p| < \delta .\] If \(f(p) > 0\), the left inequality gives \(f(x) > 0\). If \(f(p) < 0\), the right inequality gives \(f(x) < 0\). □

Proof. Suppose \(f\) is continuous at \(a\) with \(f(a)>0\); the case \(f(a)<0\) follows by applying this to \(-f\). Take \(\varepsilon = \frac {f(a)}{2} > 0\) in the definition of continuity: there is \(\delta >0\) such that \[|f(x)-f(a)| < \frac {f(a)}{2} \qquad \text {whenever } x\in [a,b] \text { and } |x-a|<\delta .\] Then \(f(x) > f(a) - \frac {f(a)}{2} = \frac {f(a)}{2} > 0\) for \(x\in [a,a+\delta )\). The argument at \(b\) is the same. □

Proof. Let \(A = \{ x\in [a,b]: f(x) < 0\}\). Then \(a\in A\), so \(A\) is non-empty, and \(A\) is bounded above by \(b\). By the completeness axiom \(A\) has a supremum; let \(p = \sup A\). Then \(a\leq p\leq b\).

Suppose \(f(p) < 0\). Then \(p\neq b\), since \(f(b)>0\), and by the sign-preservation theorems there is a \(\delta >0\) with \(f(x)<0\) for all \(x\in [p,p+\delta )\cap [a,b]\). Such points beyond \(p\) lie in \(A\), contradicting \(p = \sup A\).

Suppose \(f(p) > 0\). Then \(p\neq a\), since \(f(a)<0\), and there is a \(\delta >0\) with \(f(x)>0\) for all \(x\in (p-\delta ,p]\cap [a,b]\). So no point of \((p-\delta , p]\) lies in \(A\), and since no point of \(A\) exceeds \(p\), \(p-\delta \) is an upper bound of \(A\), again contradicting \(p = \sup A\).

Hence \(f(p) = 0\), and \(p\neq a,b\) because \(f(a)\neq 0\neq f(b)\). □

Proof. Suppose \(f(a) < f(b)\) and let \(f(a) < K < f(b)\). (If \(f(a)>f(b)\), apply the argument to \(-f\); if \(f(a)=f(b)\) there is nothing to prove.) The function \(g(x) = f(x) - K\) is continuous on \([a,b]\), with \[g(a) = f(a) - K < 0 \qquad \text {and}\qquad g(b) = f(b) - K > 0 .\] By the previous theorem there exists \(p\in (a,b)\) such that \(g(p) = 0\), that is, \(f(p) = K\). □

Proof. Let \(I\) be an interval, \(f\) continuous on \(I\), and let \(u<w<v\) with \(u,v\in f(I)\). Say \(u=f(c)\) and \(v=f(d)\) with \(c,d\in I\). Since \(I\) is an interval it contains the closed interval with end points \(c\) and \(d\), and \(f\) is continuous there, so by the Intermediate Value Theorem \(f\) takes the value \(w\) at some point between \(c\) and \(d\), which is a point of \(I\). Hence \(w\in f(I)\).

So \(f(I)\) contains every point between any two of its points, which means that it is an interval. □

Proof. There exist points \(\alpha , \beta \in [a,b]\) with \(f(\alpha ) = \sup \{f(x): x\in [a,b]\}\) and \(f(\beta ) = \inf \{f(x): x \in [a,b]\}\). Applying the Intermediate Value Theorem to \(f\) on the closed interval with end points \(\alpha \) and \(\beta \), we see that \(f\) assumes every value between \(f(\beta )\) and \(f(\alpha )\). □

Solution. Let \(f(x) = x^5-3x+1\), which is continuous. Since \(f(0) = 1>0\) and \(f(1) = -1<0\), the Intermediate Value Theorem gives a root in \((0,1)\). Bisecting:

So the root lies in \((0.3125,0.375)\), an interval of length \(0.0625\). Each bisection halves the interval, so \(n\) steps locate the root within \(2^{-n}\); continuing gives the root \(0.3347\) to four decimal places.

Solution. Let \(g(x) = \cos x - x\), continuous on \(\left [0,\frac {\pi }{2}\right ]\). Then \(g(0) = 1>0\) and \(g\left (\frac {\pi }{2}\right ) = -\frac {\pi }{2}<0\), so by the Intermediate Value Theorem \(g\) has a zero there. It has only one, because \(g\) is strictly decreasing: if \(x<y\) then \(\cos x\geq \cos y\) on this interval and \(-x>-y\), so \(g(x)>g(y)\). (The solution is approximately \(0.739\).)

Solution. The function is continuous on the closed bounded interval \([-2,2]\), so it attains a maximum and a minimum. Since \(f(x) - 2 = (x+1)^2(x-2)\leq 0\) and \(f(x)+2 = (x-1)^2(x+2)\geq 0\) on \([-2,2]\), we have \(-2\leq f(x)\leq 2\), with \(f = 2\) at \(x = -1, 2\) and \(f = -2\) at \(x = -2, 1\). So the maximum is \(2\) and the minimum is \(-2\), and by the corollary to the Intermediate Value Theorem \(f([-2,2]) = [-2,2]\).