14.5 Inverse Function Theorems

14.38   Lemma. Let $f$ be a strictly increasing continuous function whose domain is an interval $[a,b]$. Then the image of $f$ is the interval $[f(a),f(b)]$, and the function $f:[a,b] \to [f(a),f(b)]$ has an inverse.

Proof: It is clear that $f(a)$ and $f(b)$ are in image$(f)$. Since $f$ is continuous we can apply the intermediate value property to conclude that for every number $z$ between $f(a)$ and $f(b)$ there is a number $c \in [a,b]$ such that $z = f(c)$, i.e. $[f(a),f(b)] \subset$ image$(f)$. Since $f$ is increasing on $[a,b]$ we have $f(a) \leq f(t) \leq f(b)$ whenever $a \leq t \leq b$, and thus image $(f) \subset [f(a),f(b)]$. It follows that $f:[a,b] \to [f(a),f(b)]$ is surjective, and since strictly increasing functions are injective, $f$ is bijective. By remark (14.22) $f$ has an inverse.

14.39   Exercise. State and prove the analogue of lemma 14.38 for strictly decreasing functions.

14.40   Exercise. Let $f$ be a function whose domain is an interval $[a,b]$, and whose image is an interval. Does it follow that $f$ is continuous?

14.41   Exercise. A Let $f$ be a continuous function on a closed bounded interval $[a,b]$. Show that the image of $f$ is a closed bounded interval $[A,B]$.

14.42   Exercise. Let $J$ and $I$ be non-empty intervals and let $f :J \to I$ be a continuous function such that $I=$ image($f$).

a) Show that if $f$ is strictly increasing, then the inverse function for $f$ is also strictly increasing.

b) Show that if $f$ is strictly decreasing, then the inverse function for $f$ is also strictly decreasing.

14.43   Theorem (Inverse function theorem.) Let $f$ be a continuous
strictly increasing function on an interval $J=[a,b]$ of positive length, such that $f'(x) > 0$ for all $x \in \mbox{{\em interior}}(J)$. Let $I$ be the image of $J$ and let

$$g: I \to J$$

be the inverse function for $f$. Then $g$ is differentiable on the interior of $I$ and

$$g'(s) = \frac{1}{f'(g(s))} \mbox{ for all }s \in {\rm interior}(I)$$ (14.44)

Remark: If $l$ is a nonvertical line joining two points $(p,q)$ and $(r,s)$ then the slope of $l$ is

$$m = \frac{s-q}{r-p}.$$

The reflection of $l$ about the line whose equation is $y=x$ passes through the points $(q,p)$ and $(s,r)$, so the slope of the reflected line is

$$\frac{r-p}{s-q} = \frac{1}{m}.$$

Thus theorem 14.43 says that the tangent to graph($g$) at the point $(s,g(s))$ is obtained by reflecting the tangent to graph($f$) at $(g(s),s)$ about the line whose equation is $y=x$. This is what you should expect from the geometry of the situation.

Proof of theorem 14.43:      The first thing that should be done, is to prove that $g$ is continuous. I am going to omit that proof and just assume the continuity of $g$, and then show that $g$ is differentiable, and that $g'$ is given by formula (14.44).

Let $s$ be a point in the interior of $\mbox{{\rm dom}}{(g)}$. then

$$\lim_{t \to s} \frac{g(t)-g(s)}{t-s} = \lim_{t \to s} \frac{g(t)-g(s)} {f(g(t)) - f(g(s))} \mbox{{}}$$

$$= \lim_{t \to s} \frac{1}{\frac{f(g(t))-f(g(s))}{g(t)-g(s)}}.$$ (14.45)

(Observe that we have not divided by zero). Let $\{t_n\}$ be a sequence in $\mbox{{\rm dom}}{(g)}\setminus\{s\}$ such that $\{t_n\} \to s$. Then $\{g(t_n)\} \to g(s)$ (since $g$ is assumed to be continuous), and $g(t_n) \neq g(s)$ for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ (since $g$ is injective). Since $f$ is differentiable at $g(s)$, it follows that

$$\left\{ \frac{f(g(t_n)) - f(g(s))}{g(t_n) - g(s)}\right \} \to f'(g(s)).$$

Since $f'(g(s)) \not = 0$ it follows that

$$\left \{ \frac{1}{\frac{f(g(t_n)) - f(g(s))}{g(t_n) - g(s)} }\right\} \to \frac{1}{f'(g(s)}.$$

It follows that

$$\lim_{t \to s} \frac{g(t)-g(s)}{t-s} = \frac{1}{f'(g(s))}$$

and the theorem is proved. $\diamondsuit$

Remark: The inverse function theorem also applies to continuous functions $f$ on $J$ such that $f'(s) < 0$ for all $\s$ in interior $(a,b)$. Formula (14.44) is valid in this case also.

Remark: Although we have stated the inverse function theorem for functions on intervals of the form $[a,b]$, it holds for functions defined on any interval. Let $J$ be an interval, and let $f$ be a continuous strictly increasing function from $J$ to R such that $f'(x) > 0$ for all $x$ in the interior of $J$. Let $p$ be a point in the interior of image$(J)$. Then we can find points $r$ and $s$ in image$(J)$ such that $r < p < s$. Now $f$ maps the interval $[g(r),g(s)]$ bijectively onto $[r,s]$, and since $p\in (r,s)$ we can apply the inverse function theorem on the interval $[g(r),g(s)]$ to conclude that $${ g'(p) = {1\over f'(g(p))}}$$. It is not necessary to remember the formula for $g'(p)$. Once we know that $g$ is differentiable, we can calculate $g'$ by using the chain rule, as illustrated by the examples in the next section.