The value of $x$, for which the function $f(x)=x(x-2)$ attains its minimum value, is:

$\frac{1}{2}$ 1 2 0

1

$f(x) =x(x-2)$ $=x^2-2 x$ so $f'(x)=2 x-2$ differential w.r.t x finding critical points as f'(x) = 0 ⇒ 2x - 2 = 0 ⇒ x = 1 f''(x) = 2  differentiating f'(x) w.r.t x for f''(1) = 2 > 0 ⇒ 1 is point of minima