Let `f:[4,oo)to[4,oo)` be defined by `f(x)=5^(x^((x-4)))`.Then `f^(-1)(x)` isA. `2-sqrt(4-logs x)`B. `2+sqrt(4+logs x)`C. `((1)/5)^(x^(x+4))`D. not defined

Let `f:[4,oo)to[4,oo)` be defined by `f(x)=5^(x^((x-4)))`.Then `f^(-1)

1.	Let `f:[4,oo)to[4,oo)` be defined by `f(x)=5^(x^((x-4)))`.Then `f^(-1)(x)` isA. `2-sqrt(4-logs x)`B. `2+sqrt(4+logs x)`C. `((1)/5)^(x^(x+4))`D. not defined
Answer» Correct Answer - B Clearly, `f:[4,oo)to[4,oo)` is a bijection. So, its is invertible. f(x)=y, Then `5^(x^(x-4))=y` `Rightarrowx^(2)-4x=log_(5)y` `Rightarrow x^(2)-4x-log_(5)y=0` `Rightarrow x=(4pmsqrt16+4log_(5)y)/(2)` `Rightarrow f^(-1)(y)=2+sqrt(4+log_(5)y)` `Rightarrow f^(-1)(y)=2+sqrt(4+log_(5)y) " "[thereforex ge 4]` Hence, `f^(-1)(x)=2+sqrt(4+log_(5)x)` We know that if (g) x is inverse of a bijection f(x) then `fog(x)=x Rightarrowf(g(x))=x` This relation suggests the following method for finding the inverset of a bijection.