Question

Let f(x) = (x \[-\] a)² + (x \[-\] b)² + (x \[-\] c)². Then, f(x) has a minimum at x = _____________ .

पर्याय

• \[\frac{a + b + c}{3}\]

• \[\sqrt[3]{abc}\]

• \[\frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}\]

• none of these

Answer 1

\[\frac{a + b + c}{3}\]

 

\[\text { Given }:   f\left( x \right) =    \left( x - a \right)^2  +  \left( x - b \right)^2  +  \left( x - c \right)^2 \] 

\[ \Rightarrow f'\left( x \right) =   2\left( x - a \right) + 2\left( x - b \right) + 2\left( x - c \right)\] 

\[\text { For  a  local  maxima  or  a  local  minima, we  must  have }  \] 

\[f'\left( x \right) = 0\] 

\[ \Rightarrow 2\left( x - a \right) + 2\left( x - b \right) + 2\left( x - c \right) = 0\] 

\[ \Rightarrow 2x - 2a + 2x - 2b + 2x - 2c = 0\] 

\[ \Rightarrow 6x = 2\left( a + b + c \right)\] 

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

\[\text { Now },   \] 

\[f''\left( x \right) = 2 + 2 + 2 = 6 > 0\] 

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