Question

Find the values of x for which $y=[x(x-2)]^2$ is an increasing function.

Answer 1

We have,

y = [x(x-2)]2 = [x2-2x]2

$\frac {dy}{dx}$ = y'(x2-2x) = 4x(x-2)(x-1)

$\frac {dy}{dx}$ = 0$\implies$x=0, x=2, x=1.

The points x = 0, x = 1, and x = 2 divide the real line into four disjoint intervals i.e.,

(-∞,0),(0,1)(1,2), and (2,∞).

In intervals (-∞,0) and (1,2), $\frac {dy}{dx}$<0

∴ y is strictly decreasing in intervals (-∞,0) and (1,2).

However, in intervals (0, 1) and (2, ∞), $\frac {dy}{dx}$>0.

∴ y is strictly increasing in intervals (0,1) and (2,∞).

y is strictly increasing for 0<x<1 and x>2.