Let [tex]f(x) = \begin{cases} -1, & -2 \le x < 0 \\ x^2 - 1, & 0 \le x \le 2 \end{cases}[/tex] and [tex]g(x) = |f(x)| + f(x)[/tex]. Then, in the interval (-2, 2), g is:
(1) differentiable at all points
(2) not differentiable at two points
(3) not continuous
(4) not differentiable at one point
Answer (1)
To analyze the function g ( x ) = ∣ f ( x ) ∣ + f ( x ) , where f ( x ) = { − 1 , x 2 − 1 , − 2 ≤ x < 0 0 ≤ x ≤ 2 , we need to examine the differentiability and continuity of g ( x ) over the interval ( − 2 , 2 ) .
Step 1: Understanding f ( x ) and ∣ f ( x ) ∣
For − 2 ≤ x < 0 , f ( x ) = − 1 . Thus, ∣ f ( x ) ∣ = 1 because the absolute value of − 1 is 1 .
For 0 ≤ x ≤ 2 , f ( x ) = x 2 − 1 . Here, ∣ f ( x ) ∣ is the same as f ( x ) because x 2 − 1 is non-negative for 0 ≤ x ≤ 2 .
Step 2: Calculate g ( x )
For − 2 ≤ x < 0 , g ( x ) = ∣ f ( x ) ∣ + f ( x ) = 1 − 1 = 0 .
For 0 ≤ x ≤ 2 , g ( x ) = ∣ f ( x ) ∣ + f ( x ) = ( x 2 − 1 ) + ( x 2 − 1 ) = 2 ( x 2 − 1 ) .
Step 3: Analyzing Continuity and Differentiability
Continuity at x = 0 :
As x → 0 − : g ( x ) = 0 .
As x → 0 + : g ( x ) = 2 ( 0 2 − 1 ) = − 2 .
Since the left-hand limit and right-hand limit are different, g ( x ) is not continuous at x = 0 .
Because g ( x ) is not continuous at x = 0 , it cannot be differentiable at this point.
Conclusion
Based on the analysis, the best choice from the options given is (3) not continuous .
Thus, the correct multiple choice answer is (3).
