If the solve the problem

Question:

Let $f(x)=(x-a)^{2}+(x-b)^{2}+(x-c)^{2}$. Then, $f(x)$ has a minimum at $x=$

(a) $\frac{a+b+c}{3}$

(b) $\sqrt[3]{a b c}$

(c) $\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$

(d) none of these

Solution:

(a) $\frac{a+b+c}{3}$

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

$f^{\prime}(x)=0$

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

$\Rightarrow 2 x-2 a+2 x-2 b+2 x-2 c=0$

$\Rightarrow 6 x=2(a+b+c)$

$\Rightarrow x=\frac{a+b+c}{3}$

Now,

$f^{\prime \prime}(x)=2+2+2=6>0$

So, $x=\frac{a+b+c}{3}$ is a local minima.

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