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) $\sqrt[3]{abc}$

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

(d) none of these

Answer 1

$$\begin{gathered} \left(\mathrm{a}\right)\frac{a+b+c}{3} \\ \mathrm{Given}:f\left(x\right)={\left(x-a\right)}^2+{\left(x-b\right)}^2+{\left(x-c\right)}^2 \\ \Rightarrow f\text{'}\left(x\right)=2\left(x-a\right)+2\left(x-b\right)+2\left(x-c\right) \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 2\left(x-a\right)+2\left(x-b\right)+2\left(x-c\right)=0 \\ \Rightarrow 2x-2a+2x-2b+2x-2c=0 \\ \Rightarrow 6x=2\left(a+b+c\right) \\ \Rightarrow x=\frac{a+b+c}{3} \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=2+2+2=6>0 \\ \mathrm{So},x=\frac{a+b+c}{3}\mathrm{is}\mathrm{a}\mathrm{local}\text{minima.} \end{gathered}$$