# Find x which minimizes `f(x)=x - sqrt(1-x^2)`

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### 1 Answer

Given `f(x)=x-sqrt(1-x^2)`

`f'(x)=1-1/2(1-x^2)^(-1/2)*(-2x)`

`=1+x/sqrt(1-x^2)`

At maximum or minimum,`f'(x)=0`

so, `1+x/sqrt(1-x^2)=0`

`rArr x/sqrt(1-x^2)=-1`

`rArrx=-sqrt(1-x^2)`

squaring both sides, we get

`x^2=1-x^2`

`rArr 2x^2=1`

`rArr x^2=1/2`

`rArr x=+-1/sqrt2`

When, `x=1/sqrt2`

`f(1/sqrt2)=1/sqrt2-sqrt(1-1/2)`

`=1/sqrt2-1/sqrt2`

`=0`

When`x=-1/sqrt2`

`f(-1/sqrt2)=-1/sqrt2-sqrt(1-1/2)`

`=-1/sqrt2-1/sqrt2`

`=-sqrt2`

Therefore, f(x) is minimum at `x=-1/sqrt2`

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omg that's so simple, i just thought too hard

thank you sir