0
$\begingroup$

The function

f (x) = 2|x| + |x + 2| - ||x + 2| - 2|x||

has local maxima or minima at :

*A) -2

B) -2/3

C) 2/3

D) 2*

I can solve the above question graphically. I want to know how can I solve it by using calculus. Can some write an answer describing the calculus method. It will be much helpful for me.

  • 0
    Ok, following my suggested approach on what intervals can the inner absolute values be simplified? – 2017-02-01

2 Answers 2

2

It's a piecewise function. On each piece it's linear. Hence on each piece, it's either constant, or strictly monotonic (in which case, the extreme values for such a piece can only occur at the endpoints). Find the places where the pieces change. For each interval between those places, find the formula for the pieces (the absolute value symbols can be removed -- with appropriate sign changes). Find the extrema on each piece.

This approach, while algebraic, is consistent with the graphical approach, and in some sense, it can be regarded as the same approach. With that understanding, the graphical approach is the Calculus approach.

2

Recall that $\dfrac{a+b-|a-b|}{2}=min(a,b)$

Therefore,$f(x)=2 min\left(2|x|,|x+2|\right)$