JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Application of Derivatives

We analyze the function piecewise by considering different intervals:

$f(x)=\{\begin{array}{ll}-2x-4,& x<-2\\ 2x+4,& -2\le x<-\frac{2}{3}\\ -4x,& -\frac{2}{3}\le x<0\\ 4x,& 0\le x<2\\ 2x+4,& x\ge 2\end{array}$

Examining the slopes:

• For $x<-2$: slope = -2
• For $-2\le x<-2/3$: slope = 2 (increasing)
• For $-2/3\le x<0$: slope = -4 (decreasing)
• For $0\le x<2$: slope = 4 (increasing)
• For $x\ge 2$: slope = 2

At $x=-2$: the function changes from decreasing to increasing, so this is a local minimum.

At $x=-2/3$: the function changes from increasing to decreasing, so this is a local maximum.

At $x=0$: the function changes from decreasing to increasing, so this is a local minimum.

Therefore, local extrema occur at $x=-2$ and $x=-2/3$.