Observe that, #f(x)=4x^2-2x+x/(x-1/4); x in RR-{1/4}.#
#=4x^2-2x+1/4-1/4+{(x-1/4)+1/4}/(x-1/4); xne1/4#
#=(2x-1/2)^2-1/4+{(x-1/4)/(x-1/4)+(1/4)/(x-1/4)}; xne1/4#
#=4(x-1/4)^2-1/4+{1+(1/4)/(x-1/4)}; xne1/4#
# :. f(x)=4(x-1/4)^2+3/4+(1/4)/(x-1/4); xne1/4.#
Now, for Local Extrema, #f'(x)=0,# and,
#f''(x) > or < 0," according as "f_(min) or f_(max)," resp."#
#f'(x)=0#
#rArr 4{2(x-1/4)}+0+1/4{(-1)/(x-1/4)^2}=0...(ast)#
#rArr 8(x-1/4)=1/{4(x-1/4)^2}, or, (x-1/4)^3=1/32=2^-5.#
# rArr x=1/4+2^(-5/3)#
Further, #(ast) rArr f''(x)=8-1/4{-2(x-1/4)^-3}," so that, "#
#f''(1/4+2^(-5/3))=8+(1/2)(2^(-5/3))^-3>0#
#"Therefore, "f_(min)=f(1/4+2^(-5/3))#
#=4(2^(-5/3))^2+3/4+(1/4)/(2^(-5/3))=2^2*2^(-10/3)+3/4+2^(-2)*2^(5/3)#
#=1/2^(4/3)+3/2^2+1/2^(1/3)=(2^(2/3)+3+2^(5/3))/4.#
Thus, #f_(min)=f(1/4+2^(-5/3))=(2^(2/3)+3+2^(5/3))/4.#
Enjoy Maths.!
