Inverse and Composite Functions

Inverse and Composite Functions
Definition (Inverse function) : 
    If \( f:A \to B\) is a bijection, then the relation \({f^{ - 1}} = \{ (b,a)/(a,b) \in f\}\) is a function from B to A and is called the Inverse of ‘f’ and also  \(f^{-1}\) is unique and bijection
Theorem : If \(f:A \to B\) is an injection, then \(f^{-1}\) is a bijection from f(A) to A \(f(A) \subseteq B\)
Proof :
     Let \(f:A \to B\) is an injection clearly \(f^{-1}\) is a relation from f(A) to A. Then there existss at least one  \(a\in A\) such that f(a)=b.  
    Thus  \(b \in f(A)(f(A) \subseteq B),\exists \) given  a unique \(a \in A\) such that \((a,b) \in f \Rightarrow (b,a) \in {f^{ - 1}}\)
    Hence \(f^{-1}\) is a function from f(A) to A and \(f^{-1}(b)=a\) if and only if f(a)=b clearly \(f^{-1}\) is a surjection.
    Let \({b_1},{b_2} \in f(A) \Rightarrow {f^{ - 1}}({b_1}) = {f^{ - 1}}({b_2}) = a\)
   \( \Rightarrow {b_1} = {b_2} = f(a)\) .  Therefore \(f^{-1}\) is an Injection
   \(\Rightarrow {f^{ - 1}}:f(A) \to A\) is a bijection
    \(\Rightarrow f:A \to f(A)\) is also bijection
Note : (1) If \(f:A \to B\) is a bijection then \({f^{ - 1}}:B \to A\) is also bijection.
    (2)   \({\left( {{f^{ - 1}}} \right)^{ - 1}} = f\) 
    (3) The inverse of a bijection is also bijection and unique.
ALGORITHM :
    Let \(f:A \to B\) be a bijection defined as f(x) =2x -1.  To find the Inverse of ‘f’.  We have the following steps
Step I : Let \(2x - 1 = y\,\,\,\,\,i.e\,\,\,y = f(x)\)
Step II : Solve to obtain x in term of y i.e .\(2x = y + 1 \Rightarrow x = \frac{{y + 1}}{2}\)
Step III : In the relation obtained in step II replace x by \({f^{ - 1}}(y)\)
    Generally we represent functions in terms of x
    i.e \({f^{ - 1}}(y) = \frac{{y + 1}}{2} \Rightarrow {f^{ - 1}}(x) = \frac{{x + 1}}{2}\)
Note : If the sets A and B will have the same number of elements then only bijection is possible from A to B.
    If \(f:A \to B\) is bijection then \(f^{-1}:B \to A\) is also bijection and unique.
INVERSE TRIGONOMETRIC FUNCTIONS
    Generally we know that in algebra bijection mapping only have Inverses.
    Trigonometric functions are not one-one in their natural domains.  But if we restrict the domains definition of thses functions to appropriate subsets, then they become bijective.
Consider f(x)=sinx whose domain is R and the range is the subset of real numbers lying between -1 and +1 \(\,\,\left( { - 1 \leqslant \sin x \leqslant 1} \right)\) .
 ‘f’ is a funtion from R to \(\{ y| - 1 \leqslant y \leqslant 1\} \) clearly ‘f’ is not one-one function, but it is an onto function.
    for \(sin x =sin{ \pi \over 3}\)
General solution \(x=n\pi+(-1)^n.\pi/3(n\in z)\)
    \(\therefore f(x) = \sin x,x \in R\) doesn’t have an Invere function
Consider the subset { \(x|-\pi/2 \le x \le\pi/2\) }of R
 For \(f(x)=sinx, f:[-\pi/2,\pi/2]\to [ - 1,1]\) is both one-one and onto i.e., bijective.
    \( \therefore \,y = \sin x \Rightarrow x = {\sin ^{ - 1}}y \Rightarrow {f^{ - 1}}(y) = {\sin ^{ - 1}}y\)
Similarly \({\cos ^{ - 1}}x,{\tan ^{ - 1}}x,{\sec ^{ - 1}}x,\cos e{c^{ - 1}}x,{\cot ^{ - 1}}x\) are also bijection which are Inverse of \(\cos x,\tan x,\sec x,\csc x\) and cot x respectively.
Inverse function of a circular function is known as an Inverse circular function