What f’(x) equal to when 0 < x < 1?

Concept:

• \( \left| {\rm{x}} \right| = \left\{ {\begin{array}{*{20}{c}} { - x,\;\;x < 0}\\ {x,\;\;x \ge 0} \end{array}} \right.\)
• \( \left| {{\rm{x}} - 1} \right| = \left\{ {\begin{array}{*{20}{c}} { - \left( {x - 1} \right),\;\;x < 1}\\ {x - 1,\;\;x \ge 1} \end{array}} \right.\)

Calculation:

f(x) = (|x| - |x – 1|) 2  when 0 < x < 1

For x > 0, |x| = x 

For x < 1, |x - 1| = -(x - 1)

∴ f(x) = (x - (- (x - 1)))2

⇒ f(x) = (x + x – 1)2

⇒ f(x) = (2x – 1)2

⇒ f(x) = 4x2 - 4x + 1

Now, f’(x) = 8x – 4

Hence, option (4) is correct.