What are the local extrema of #f(x)= 4x^2-2x+x/(x-1/4)#?

1 Answer
Apr 18, 2017

Answer:

# f_(min)=f(1/4+2^(-5/3))=(2^(2/3)+3+2^(5/3))/4.#

Explanation:

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.!