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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