Let's analyze the function g(x) = |f(x)| + f(|x|) in the interval (-2, 2), where f(x) is defined piecewise:

$f\left(x\right)=\{\begin{array}{c}-1,-2\le x<0\\ x^2-1,0\le x\le 2\end{array}$

Since g(x) involves |x|, we need to consider two cases: x ≥ 0 and x < 0.

Case 1: x ≥ 0 (0 ≤ x < 2)

Here, |x| = x. So:

f(x) = x² - 1 (since x ≥ 0)

f(|x|) = f(x) = x² - 1

|f(x)| = |x² - 1|

Thus, g(x) = |x² - 1| + (x² - 1)

Now, note that x² - 1 is negative when 0 ≤ x < 1, and non-negative when 1 ≤ x ≤ 2.

So, we break this case further:

Subcase 1a: 0 ≤ x < 1

Then x² - 1 < 0, so |x² - 1| = -(x² - 1) = 1 - x²

Therefore, g(x) = (1 - x²) + (x² - 1) = 0

Subcase 1b: 1 ≤ x ≤ 2

Then x² - 1 ≥ 0, so |x² - 1| = x² - 1

Therefore, g(x) = (x² - 1) + (x² - 1) = 2(x² - 1)

Case 2: x < 0 (-2 ≤ x < 0)

Here, |x| = -x (which is positive). So:

f(x) = -1 (since x < 0)

f(|x|) = f(-x). Since -x > 0, f(-x) = (-x)² - 1 = x² - 1

|f(x)| = |-1| = 1

Thus, g(x) = 1 + (x² - 1) = x²

Summary of g(x) in (-2,2):

$g\left(x\right)=\{\begin{array}{c}{x}^2,-2\le x<0\\ 0,0\le x<1\\ 2(x^2-1),1\le x<2\end{array}$

Continuity Check:

Let's check at the boundaries x=0 and x=1.

At x=0:

Left-hand limit (x→0⁻): g(x) = x² → 0

Right-hand limit (x→0⁺): g(x) = 0 → 0

g(0) = 0 (from right definition)

So, continuous at x=0.

At x=1:

Left-hand limit (x→1⁻): g(x) = 0 → 0

Right-hand limit (x→1⁺): g(x) = 2(x²-1) → 2(1-1)=0

g(1) = 2(1²-1)=0

So, continuous at x=1.

g(x) is continuous everywhere in (-2,2).

Differentiability Check:

Now, check differentiability at x=0 and x=1.

At x=0:

Left derivative (x<0): g'(x) = 2x → 0 as x→0⁻

Right derivative (x>0): g'(x) = 0 (for 0

So, g'(0) exists and is 0.

At x=1:

Left derivative (x<1): g'(x) = 0 → 0 as x→1⁻

Right derivative (x>1): g'(x) = 4x → 4 as x→1⁺

Left derivative (0) ≠ Right derivative (4), so not differentiable at x=1.

Also check at x=-2? But x=-2 is not in the open interval (-2,2), so we don't consider it.

Thus, in (-2,2), g(x) is not differentiable at exactly one point (x=1).

Final Answer:

g is not differentiable at one point in the interval (-2,2).

Related Topics and Formulae:

Piecewise Functions: Functions defined by different expressions on different intervals. Key to analyze each piece separately and at boundaries.

Continuity: A function f is continuous at x=a if $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)$.

Differentiability: A function is differentiable at x=a if the left-hand derivative equals the right-hand derivative. For piecewise functions, check derivatives from both sides at boundaries.

Absolute Value: |a| = a if a≥0, and -a if a<0. Important in transformations like |f(x)|.