Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 1 / 9F (Analysis I)

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.

True. If $x_n\to x$, continuity of $g$ gives $g(x_n)\to g(x)$, and continuity of $f$ then givesEquivalently, this is the composition of continuous functions theorem.

This can be false because a strictly increasing $g$ need not be surjective. TakeThen $g$ is strictly increasing and $f\circ g=0$ is continuous, while $f$ is discontinuous at zero.

True. Part (a) shows that $f$ is strictly monotone. Suppose it is increasing and let $y_0=f(x_0)$. Given $ϵ>0$,Choose $\delta >0$ smaller than both distances from $y_0$ to the two outer values. If $|y-y_0|<\delta$, monotonicity forcesThus $f^{-1}$ is continuous at $y_0$. The decreasing case is identical with inequalities reversed.

This can be false. The functionis a differentiable bijection of $\mathbb{R}$, buthas an unbounded difference quotient at zero and is not differentiable there. The inverse function theorem requires the additional local condition $f^{\prime }(x)\ne 0$.