Let `f(x)=(x-a)^2+(x-b)^2+(x-c)^2dot` Then, `f(x)` has a minimum at `x=` `(a+b+c)/3` (b) `1/2` (c) `1/8` (d) none of these

Explanation

`frac{a+b+c}{3}`



Given : `f(x)=(x-a)^{2}+(x-b)^{2}+(x-c)^{2}` `Rightarrow f^{prime}(x)=2(x-a)+2(x-b)+2(x-c)`


For a local maxima or a local minima, we must have `f^{prime}(x)=0`


`Rightarrow 2(x-a)+2(x-b)+2(x-c)=0`


`Rightarrow 2 x-2 a+2 x-2 b+2 x-2 c=0`


`Rightarrow 6 x=2(a+b+c)`


`Rightarrow x=frac{a+b+c}{3}`


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