Inverse functions

Injectivity

A function f is called injective if for all x_1,x_2 \in \mathrm{Dom}(f), \space x_1 \neq x_2 implies that f(x_1) \neq f(x_2). Meaning that for every y \in \mathrm{Rang}(f) there is precisely one x \in \mathrm{Dom}(f) such that y = f(x). Meaning, every x has an unique y.

Inverse function

If f is injective, then it has an inverse function f^{-1}. The value of f^{-1}(x) is the unique number y in the domain of f for which f(y) = x. Thus,

y = f^{-1}(x) \iff x = f(y)

Suppose f is a continuous function, f is injective if f is strictly increasing or decreasing. That is, f' \leq 0 \vee f' \geq 0.

Derivative of inverse function

When f is differentiable and injective (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}.

Proof:

f(y) = x \implies f'(y) \frac{dy}{dx} = 1

\frac{dy}{dx} = \frac{1}{f'(y)} = \frac{1}{f'(f^{-1}(x))}

Without knowing the inverse function a value of the inverse derivative may be determined.

The arcsine function

Always \arcsin not \sin^{-1} that is wrong since \sin is not injective.

For x \in [-\frac{\pi}{2},\frac{\pi}{2}] \space \arcsin(\sin x) = x

For x \in [-1,1] \space \sin(\arcsin x) = x

The arccosine function is similar.

Example question

Prove that \forall x \geq 0: \arctan(x + 1) - \arctan(x) < \frac{1}{1 + x^2}.

For x = 0: \frac{\pi}{4} < 1.

For x > 0: Consider the function f(t) = \arctan(t) on the interval [x, x+1]. Apply the Mean-value theorem of f at the interval [x,x+1],

\frac{f(x+1) - f(x)}{(x+1) - 1} = f'(c).

Let \arctan(c) = y then, c = \tan y,

\begin{array}{ll} \frac{dy}{dc} (c = \tan y) &\implies 1 = \sec^2 (y) \frac{dy}{dc} = (\tan^2 y + 1) \frac{dy}{dc} \ &\implies 1 = (c^2 + 1) \frac{dy}{dc} \ &\implies \frac{dy}{dc} = \frac{1}{c^2 + 1}. \end{array}

Obtaining,

\arctan(x+1) - \arctan(x) = f'(c) = \frac{1}{c^2 + 1}.

For some c \in (x,x+1), since c > x

\frac{1}{1 + c^2} < \frac{1}{1 + x^2},

thereby

\arctan(x+1) - \arctan(x) < \frac{1}{1 + x^2}.

2.1 KiB Executable file Raw Permalink Blame History

Inverse functions

Injectivity