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$y=x(x-3)^2$ decreases for the value of x given by |
$0<x<\frac{3}{2}$ x > 0 1 < x < 3 x < 0 |
1 < x < 3 |
The correct answer is Option (3) - $1 < x < 3$ $y=x(x-3)^2$ $y'=(x-3)^2+2x(x-3)$ $=(x-3)(3x-3)=0$ $x=1,3$
Y is decreasing for $1 < x < 3$ |
