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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}.