Given function
#f(x)=\frac{(x-2)(x+3)}{x^2}=\frac{x^2+x-6}{x^2}#
#f'(x)=\frac{d}{dx}(\frac{x^2+x-6}{x^2})#
#=\frac{x^2\frac{d}{dx}(x^2+x-6)-(x^2+x-6)\frac{d}{dx}x^2}{(x^2)^2}#
#=\frac{x^2(2x+1)-(x^2+x-6)2x}{x^4}#
#=\frac{-x^2+12x}{x^4}#
#=\frac{-x+12}{x^3}#
Now, at the critical first derivative of given function #f(x)# must be zero i.e. #f'(x)=0#
#\therefore \frac{-x+12}{x^3}=0#
#-x+12=0\quad (\forall \ x\ne 0)#
#x=12#
Taking second derivative of given function as follows
#f''(x)=\frac{d}{dx}f(x)=\frac{d}{dx}(\frac{-x+12}{x^3})#
#f''(x)=\frac{2x-36}{x^4}#
#f''(12)=\frac{2(12)-36}{12^4}<0#
hence, the function is maximum at #x=12#
Now, the maximum value of #f(x)#
#f(12)=\frac{12^2+12-6}{12^2}#
#=25/24#
