F(x) = 3x+2, x≤2 f(x) = -0,5x-3, -2

To find the domain (D) and range (E) of a function, we need to consider the restrictions on the variable x and the possible values of the function f(x) for those values of x.

D(f): The domain of f(x) is the set of all possible values of x for which the function is defined.

In this case: For the first piece of the function, f(x) is defined for x ≤ 2. For the second piece of the function, f(x) is defined for -2 < x < 0. For the third piece of the function, f(x) is defined for x ≥ 0.

Combining all these conditions, we get the domain D(f) as: D(f) = {x | x ≤ 2 or -2 < x < 0 or x ≥ 0}

E(f): The range of f(x) is the set of all possible values that the function can take.

Looking at the given function, we can see that all three pieces of the function give constant values.

For the first piece (x ≤ 2), f(x) = 3x + 2. As x varies from negative infinity to 2 (inclusive), the function outputs all values greater than or equal to f(-∞) = -∞ and less than or equal to f(2) = 8. Therefore, the range for this piece is [-∞, 8].

For the second piece (-2 < x < 0), f(x) = -0.5x - 3. As x varies from -2 to 0 (both exclusive), the function outputs all values greater than or equal to f(-2) = -2 and less than or equal to f(0) = -3. Therefore, the range for this piece is [-3, -2].

For the third piece (x ≥ 0), f(x) = -5. Since this piece is a horizontal line, the function outputs the constant value f(x) = -5 for all x greater than or equal to 0. Thus, the range for this piece is [-5,-5].

Combining all these ranges, we get the range E(f) as: E(f) = [-∞, 8] ∪ [-3, -2] ∪ [-5, -5]

Zeros of the function: To find the zeros of the function, we need to solve for x when f(x) = 0.

For the first piece (3x + 2), when f(x) = 0, we have: 3x + 2 = 0 3x = -2 x = -2/3

For the second piece (-0.5x - 3), when f(x) = 0, we have: -0.5x - 3 = 0 -0.5x = 3 x = -6

For the third piece (-5), since it is a constant, there are no zeros.

Therefore, the zeros of the function are x = -2/3 and x = -6.

Intersection with the y-axis (Oy): To find the intersection with the y-axis, we need to find the value of f(x) when x = 0.

For the third piece (x ≥ 0), f(x) = -5. So, when x = 0, we have f(x) = -5.

Therefore, the intersection with the y-axis is at the point (0, -5).

Increasing or decreasing function: To determine if the function is increasing or decreasing, we need to examine the slope of the function in each interval.

For the first piece (3x + 2, x ≤ 2), the slope is positive (3), indicating that the function is increasing.

For the second piece (-0.5x - 3, -2 < x < 0), the slope is negative (-0.5), indicating that the function is decreasing.

For the third piece (-5, x ≥ 0), the slope is zero (since it is a horizontal line), indicating that the function is neither increasing nor decreasing.

To summarize: D(f) = {x | x ≤ 2 or -2 < x < 0 or x ≥ 0} E(f) = [-∞, 8] ∪ [-3, -2] ∪ [-5, -5] Zeros of the function: x = -2/3 and x = -6 Intersection with Oy: (0, -5) Increasing or decreasing: Increasing on x ≤ 2, Decreasing on -2 < x < 0, Neither on x ≥ 0.