Question

Given that the function $f\left(x\right)={(x-p)}^2+{(x-q)}^2+{(x-r)}^2$ has a minimum, find the corresponding values of $x$

• A. $\frac{1}{3}(p+q+r)$
• B. $\frac{2}{3}(p+q+r)$
• C. $\frac{1}{2}(p+q+r)$
• D. None of these

Answer 1

Answer: (A)

$\frac{1}{3}(p+q+r)$

We have $f\left(x\right)={(x-p)}^2+{(x-q)}^2+{(x-r)}^2$ ...(1)

It is differentiable function for $x\in (-\infty ,\infty )$.

Hence by Fermat theorem, a minimum can be attained when $f^{\prime }\left(x\right)=0$

Differentiating (1) w.r.t $x$, we get

$f^{\prime }\left(x\right)=2(x-p)+2(x-q)+2(x-r)$ ..(2)

$f^{\prime }\left(x\right)=0$ when $2(x-p)+2(x-q)+2(x-r)=0$

$\Rightarrow 3x-(p+q+r)=0\Rightarrow x=\frac{1}{3}(p+q+r)$

Since a single value of $x$ is obtained, without further investigation we can say the minimum is attained at $x=\frac{1}{3}(p+q+r)$