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

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