A. f(x) = 2|2| is differentiable overf
X
B. g(x) = 2 + || is differentiable over
-f

Recall the definition of absolute value:

• If x ≥ 0, then |x| = x

• If x < 0, then |x| = -x

(a) Splitting up f(x) = x |x| into similar cases, you have

• f(x) = x ² if x ≥ 0

• f(x) = -x ² if x < 0

Differentiating f, you get

• f '(x) = 2x if x > 0 (note the strict inequality now)

• f '(x) = -2x if x < 0

To get the derivative at x = 0, notice that f '(x) approaches 0 from either side, so f '(x) = 0 if x = 0.

The derivative exists on its entire domain, so f(x) is differentiable everywhere, i.e. over the interval (-∞, ∞).

(b) Similarly splitting up g(x) = x + |x| gives

• g(x) = 2x if x ≥ 0

• g(x) = 0 if x < 0

Differentiating gives

• g'(x) = 2 if x > 0

• g'(x) = 0 if x < 0

but this time the limits of g'(x) as x approaches 0 from either side do not match (the limit from the left is 0 while the limit from the right is 2), so g(x) is differentiable everywhere except x = 0, i.e. over the interval (-∞, 0) ∪ (0, ∞).

the axial of symmetry

step-by-step explanation:

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )