Inverse Functions

Definition. A function is called one-to-one if $f(a)\neq f(b)$ whenever $a\neq b.$

Definition. For any function $f$, the inverse of $f$, denoted by $f^{-1}$, is the set of all pairs $(a,b)$ for which $(b,a)$ is in $f.$

Theorem. $f^{-1}$ is a function if and only if $f$ is one-to-one.

Proof. (1) Assume $f$ is one-to-one. Let $(a,b)$, $(a,c)$ be two pairs in $f^{-1}$. Then $(b,a)$ and $(c,a)$ are in $f$. This means $f(b)=a=f(c)$. Since $f$ is one-to-one, it follows that $b=c$. Hence $f^{-1}$ is a function.

(2) Assume that $f^{-1}$ is a function. If $f(b) = f(c) = a$, then $f$ contains the pair $(b,a)$ and $(c,a)$. Therefore $(a,b)$ and $(a,c)$ are in $f^{-1}$. Since $f^{-1}$ is a function, this implies $b = c$. Hence $f$ is one-to-one.

Theorem. If $f$ is continuous and one-to-one on an interval, then $f$ is either increasing or decreasing on that interval.

Proof. If $f$ is neither increasing nor decreasing, then there exist $a<b$ and $c<d$ on the interval with $f(a)<f(b)$ and $f(c)>f(d)$. Consider the function $g$ defined by
MATH

Then $g$ is continuous on $[0,1]$ with $g(0)=f(a)-f(b)<0$ and $g(1)=f(c)-f(d)>0$. It follows from the Intermediate Value Theorem that there exists $s\in [0,1]$ such that $g(s)=0$. This implies MATH since $f$ is one-to-one. But the inequality
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shows that MATH, a contradiction.

Theorem. If $f$ is continuous and one-to-one on an interval, then $f^{-1}$ is also continuous.

Proof. From the preceding theorem, we see that $f$ is either increasing or decreasing on that interval. By replacing $f\ by-f$, we may assume that $f$ is increasing. Fix $b$ in the domain of $f^{-1}$ and let $\epsilon >0$. Such a number $b$ must be of the form $f(a)$ for some $a$ in the domain of $f$. We wish to find $\delta >0$ so that the inequality
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implies the inequality
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Now let $\delta $ be the smaller of MATH and MATH. This choice of $\delta $ ensures that
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Consequently, the inequality
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implies the inequality
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Since $f^{-1}$ is also increasing, we have
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i.e.,
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Theorem. If $f$ is a continuous one-to-one function defined on an interval and MATH then $f^{-1}$ is not differentiable at $a$.

Proof. Since $f(f^{-1}(x))=x$, the differentiability of $f^{-1}$ at $a$ would imply
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a contradiction.

Theorem. Let $f$ a continuous one-to-one function defined on an interval, and suppose that $f$ is differentiable at $f^{-1}(b)$, with derivative MATH. Then $f^{-1}$ is differentiable at $b$, and
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Proof. Let $b=f(a)$. Every number $b+h$ in the domain of $f^{-1}$ can be written in the form
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for some unique $k$. From this we have
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or
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The previous theorem shows that $f^{-1}$ is continuous at $b$. Therefore
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i.e., MATH

Since
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it follows from
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that
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Trigonometric Functions

Definition. MATH

Definition. For $x\in [-1,1]$, define
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It follows from the Fundamental Theorem of Calculus that $A$ is differentiable on $(-1,1)$ and
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Therefore $A$ decreases from $A(-1)=\frac \pi 2$ to $A(1)=0.$

Definition. If $0\leq x\leq \pi $, then $\cos \ x$ is the unique number in $[-1,1]$ such that
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Definition. MATH

Theorem. If $0<x<\pi $, then
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MATH

Proof. Let $B(x)=2A(x)$. Then $B(\cos \ x)=x$. Since $A$ is differentiable with MATH, it follows that $B$ is differentiable with MATH. Hence
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Therefore MATH. Since MATH we have
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For $x\in [\pi ,2\pi ]$, define
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For any real number $x$, write
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where $k$ and an integer so that MATH. Define
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Then
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Other standard trigonometric functions are defined as:
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MATH

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Theorem. If $x\neq k\pi +\pi /2$ then
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If $x\neq k\pi $, then
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Inverse Trigonometric Functions

The inverse of the function
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is denoted by $\sin {}^{-1}$ or arcsin. Thus $\sin {}^{-1}x$ is the unique number in $[-\pi /2,\pi /2]$ satisfying
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The domain of $\sin {}^{-1}$ is $[-1,1]$. The inverse of
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is denoted by $\cos {}^{-1}$ or arccos. The domain of $\cos {}^{-1}$ is $[-1,1].$ The $\cos {}^{-1}x$ is the unique number in $[0,\pi ]$ satisfying
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The inverse of the function
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is denoted by $\tan {}^{-1}$ or arctan. The domain of $\tan {}^{-1}$ is $\QTR{bf}{R}$. The $\tan {}^{-1}x$ is the unique number in $[-\pi /2,\pi /2]$ satisfying
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Theorem. For $-1<x<1$, we have
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For $x\in \QTR{bf}{R}$, we have
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Theorem. The only solution to the differential equation
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MATH
is $f\equiv 0.$

Proof. The given condition implies
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Therefore
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so MATH is constant. From MATH it follows that this constant is zero. Thus
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This implies $f\equiv 0.$

Theorem. If $f$ has second derivative everywhere and
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MATH

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then
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Proof. Let
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Then
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Consequently,
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Therefore $g\equiv 0.$

Theorem.
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Proof. Fix $y$. Define $f(x)=\sin (x+y)$. Then
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MATH
Therefore
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MATH

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The previous theorem shows
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