Inverse Functions and the Exponential

Consider the following question: Given $y=f(x)$ for $a\leqq x\leqq b$, when can we turn this relationship around, i.e., when does this mean the same thing as $x=g(y)$ for some function $g(y)$? Thus we want to have $y=f(x)$ if and only if $x=g(y)$. Observe that our notation means that $g(y)$ is to assign to each $y$ of a certain domain exactly one value of $x$. Therefore if we have $y_0=f(x_1)$, $y_0=f(x_2)$, where $x_1\ne x_2$, we cannot define such a function $g(y)$ at $y_0$. So we must have $f(x_1)\ne f(x_2)$ if $x_1\ne x_2$. We already know two important classes of functions which have this property: those which are strictly increasing and those which are strictly decreasing.

The function whose graph is shown below is strictly increasing for $0\leqq x\leqq 1$. The function $g(y)$, which we call the inverse function, is given by

$$g(y)=y-1,3/2\leqq y\leqq 2.$$

We prefer, however, to restrict our attention to functions which can be defined throughout an entire interval, rather than on a broken interval as in this case. We could eliminate the "jump" in $f(x)$ by requiring that it be continuous. This requirement gives us considerable information about the inverse function, because of the following theorem which will be proved later:

Now let $f(x)$ be continuous and strictly increasing for $a\leqq x\leqq b$. (We could also assume $f(x)$ continuous and strictly decreasing.) Then , and if $a\leqq x\leqq b$, then $f(a)\leqq f(x)\leqq f(b)$. Moreover, if $f(a)\leqq y\leqq f(b)$, there is an $x,a\leqq x\leqq b$, such that $y=f(x)$. We can then define the inverse function $g(y)$ on the whole interval $f(a)\leqq y\leqq f(b)$:

$x=g(y)$ is to mean simply that $y=f(x)$.

Once we have given a precise definition of continuity, it will be easy to see that $g(y)$ is continuous. It is clearly strictly increasing.

For the present, let us assume that $f(x)$ is also differentiable; we wish to express g'(y) in terms of f'(x). Let us also make the somewhat stronger assumption that f'(x) > 0 for $a\leqq x\leqq b$. (For instance, $y=x^3$ is strictly increasing for $-1\leqq x\leqq 1$, but y' = 0 at $0$. However, we have seen that if f'(x) > 0 for all $x$ of an interval, then $f(x)$ is strictly increasing there.) Then let , and consider

$$\frac{g(y+h)-g(y)}{h}\cdot$$

Now $y=f(x)$ and $y+h=f(x+k)$ for some $x$ and $x+k,a\leqq x\leqq b$, $a\leqq x+k\leqq b$. Therefore the difference quotient for $g(y)$ is

$$\frac{(x+k)-x}{f(x+k)-f(x)}=\frac{k}{f(x+k)-f(x)}=\frac{1}{\frac{f(x+k)-f(x)}{k}}\cdot$$

By our remark that $g(y)$ is continuous, the quantity $g(y+h)-g(y)=k$ goes to $0$ as $h\to 0$. But as $k\to 0$,

$$\frac{f(x+k)-f(x)}{k}$$

goes to f'(x). Therefore we have

g'(y) = \lim_{h \to 0} \frac{g(y+h) - g(y)}{h} = \frac{1}{f'(x)},

where $y=f(x),x=g(y)$. Thus we have the rule g'(y) = \frac{1}{f'(x)} for differentiating inverse functions.

We have seen that for is continuous and strictly increasing; moreover, its derivative for . Therefore it has an inverse function, which we denote temporarily by $E(y)$ and later by $e^y$, after we have shown that its properties entitle it to this notation. Thus we define $E(y)$, for any number $y$, by

$x=E(y)$ if and only if $y=\log x$.

If $y=\log x$, then

E'(y) = \frac{1}{(\log x)'} = \frac{1}{1/x} = x = E(y),

or E'(y) = E(y). We have already attributed this property to $e^y$. We also see that for all $y$, and that $E(0)=1$. When $y$ is far negative, $E(y)$ is close to $0$, and when $y$ is large and positive, so is $E(y)$. Now $E(y_1+y_2)$ is that number $x$ whose log is $y_1+y_2$. But if $\log x_1=y_1,\log x_2=y_2$, then $\log (x_1{x}_2)=y_1+y_2$, as we have seen. Hence the desired number is $x_1{x}_2$, i.e., $E(y_1)E(y_2)$, and we have the relation

$$E(y_1+y_2)=E(y_1)E(y_2)\cdot$$

It follows that for any positive integer $n,E(n)=E(1{)}^n$. Thus we have recovered some of the exponential-like properties of this function. Let us define a number $e$ by $e=E(1)$, so that $e$ is that number such that $\log e=1$. Then we see that for positive integers $n,E(n)=e^n$. We agree to extend this notation to all $y$, and to write $e^y=E(y)$. Thus for all $y,e^0=1,$

e^{y_1 + y_2} = e^{y_1} e^{y_2}, \quad (e^y)' = e^y \cdot

As indicated before, we can now define $a^b$ for by $a^b=e^{b\log a}$. Then

$$a^{b_1+b_2}=e^{(b_1+b_2)\log a}=e^{b_1\log a+b_2\log a}=e^{b_1\log a}{e}^{b_2\log a}=a^{b_1}{a}^{b_2},$$

and similarly we can show that

$$(a^b{)}^c=a^{bc};a^0=1;a^1=a;a^n=a\underset{n}{\underbrace{\cdot a\cdot \cdots \cdot }}a,$$

$n$ a positive integer;

$$a^{-n}=\frac{1}{a^n}\cdot$$

It is of interest to introduce some other inverse functions, notably those of the trigonometric functions. It is obvious from their periodic behavior that these functions are neither strictly increasing nor strictly decreasing; but all this means is that we cannot hope to define an inverse function whose values (whose range, in more technical language) cover the whole $x$-axis. For instance, the function $y=\sin x$ is strictly increasing for

$$-\frac{\pi }{2}\leqq x\leqq \frac{\pi }{2},$$

and y' > 0 except at the endpoints. The values run from $y=-1$ to $y=1$. Hence we can define an inverse function to $y=\sin x$, written as $x=\arcsin y$, for $-1\leqq y\leqq 1$. Its values lie between $x=-\frac{\pi }{2}$ and $x=\frac{\pi }{2}$. Its derivative is

\frac{1}{(\sin x)'} = \frac{1}{\cos x} \cdot

We should like to express this in terms of $y$. Since ${\sin }^{2}x+{\cos }^{2}x=1$, and since for , we have

$$\cos x=\sqrt{1-{\sin }^{2}x}=\sqrt{1-y^2}\cdot$$

Thus

( \arcsin y)' = \frac{1}{\sqrt{1 - y^2}}, \quad -1 < y < 1 \cdot

Likewise, $x=\arccos y$ is defined for $-1\leqq y\leqq 1$ and takes on values between $x=0$ and $x=\pi$. (Here we use the inverse function of a strictly decreasing function.)

By the above process,

( \arccos y)' = \frac{-1}{\sqrt{1 - y^2}}, \quad -1 < y < 1 \cdot

$x=\arctan y$ is similarly defined, in this case for all $y$, using the fact that $y=\tan x$ is strictly increasing for

This is then the range of values of $x=\arctan y$. As before,

( \arctan y)' = \frac{1}{1 + y^2} \cdot

As a final remark, let us add that the only functions of the form $x^a$ for which we had a really adequate definition at the beginning of this course were those for which $a$ was an integer. For $n$ a positive integer, the function $y=x^n$ is continuous and strictly increasing for $x\geqq 0$ if $n$ is even, for all $x$ if $n$ is odd. Thus we have the inverse function $x=y^{1/n}$, defined as above: $x=y^{1/n}$ if and only if $y=x^n$. Finally, we can define

$$x^{m/n}=(x^{1/n}{)}^m=(x^m{)}^{1/n},$$

for $m$ and $n$ positive; and for $b$ a negative rational number, define

$$x^b=\frac{1}{x^{-b}},x\ne 0\cdot$$

Thus $x^b$ is defined for $b$ rational; we have not yet defined $x^a$ for arbitrary $a$. This is accomplished by using $x^a=e^{a\log x}$. Since this definition coincides with the one already given for $a$ rational, we may use it as the general definition. Then we have

\begin{aligned} (x^a)' &= (e^{a \log x})' \\ &= e^{a \log x} \cdot \frac{a}{x} \\ &= e^{a \log x} \cdot a \cdot e^{-\log x} \\ &= a \cdot e^{(a-1) \log x} = a x^{a-1},\\ \end{aligned}

i.e., (x^a)' = a x^{a-1} for all $a$.