The function f ( x ) = 2 | x | + | x + 2 | - | | x + 2 | - 2 | x | | has a local minimum or a local maximum at x = [2013]

Question

The function $f (x) = 2$ $| x |$ $+$ $| x + 2 |$ $-$ $| | x + 2 | - 2 | x | |$ has a local minimum or a local maximum at $x =$ [2013]

Smart Achievers · Accepted Answer

(1, 2)

Given: $f (x) = 2$ $| x |$ $+$ $| x + 2 |$ $-$ $| | x + 2 | - 2 | x | |$

Critical points of $f (x)$ can be obtained by solving

$| x | = 0,$ $| x + 2 | = 0$ and $| | x + 2 | - 2 | x | |$ $= 0,$

which give $x = 0, - 2, 2, - \frac{2}{3}$

$∴$ $f (x) =$ ${\begin{matrix} - 2 x - 4, & x \leq - 2 \\ 2 x + 4, & - 2 < x \leq - \frac{2}{3} \\ - 4 x & - \frac{2}{3} < x \leq 0 \\ 4 x & 0 < x \leq 2 \\ 2 x + 4 & x > 2 \end{matrix}$

Graph of $y = f (x)$ is as follows:

From the graph, $f (x)$ has local minimum at $x = - 2$ and $x = 0$ and $f (x)$ has local maximum at $x = - \frac{2}{3}$

Solution

Q.
The function $f (x) = 2$ $| x |$ $+$ $| x + 2 |$ $-$ $| | x + 2 | - 2 | x | |$ has a local minimum or a local maximum at $x =$ [2013]

1 $- 2$

2 $\frac{- 2}{3}$

3 $2$

4 $\frac{2}{3}$

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Solution

Q. The function f(x)=2|x|+|x+2|-||x+2|-2|x|| has a local minimum or a local maximum at x= [2013]

1 -2

2 -23

3 2