Question

The function f(x)=x-|x-x²|,-1≤ x\le1 is

• A. [-1,1]
• B. (-1,1)
• C. [-1,1] but not differentiable at 0
• D. (-1,1)setminus0

Hint:

Absolute value functions are often continuous but fail differentiability at sign-change points.

Answer 1

The Correct Option is C

x-x² = x(1-x) f(x)= begincases 2x-x², & -1≤ x<0
x², & 0≤ x\le1 endcases Continuity: limₓtₒ₀⁻f(x)=limₓtₒ₀⁺f(x)=0=f(0) So f is continuous on [-1,1]. 
Differentiability at 0: f'_-(0)=2, f'_+(0)=0 Since derivatives are unequal, f is not differentiable at x=0.