Question

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

Answer 1

`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.!