Determine the intervals over which the function is increasing, decreasing, or constant. (Enter your answers using interval notation. If an answer does not exist, enter DNE.)

f(x) =

4x + 2, x ≤ −2
x^2 − 10, x > −2

To determine the intervals over which the function is increasing, decreasing, or constant, we need to find the intervals that satisfy each condition separately.

For the function f(x) = 4x + 2, where x ≤ -2, the function is a linear equation with a positive slope of 4. This means that the function increases for all x-values less than or equal to -2. In interval notation, this can be written as (-∞, -2].

For the function f(x) = x^2 - 10, where x > -2, the function is a quadratic equation. To determine if the function is increasing or decreasing, we can find the derivative of the function and consider its sign.

Taking the derivative of f(x), we have:

f'(x) = 2x

Since the derivative is positive when x > 0 and negative when x < 0, we can conclude that the function is increasing for x > 0 and decreasing for x < 0. However, in this case, we are only considering x > -2. Therefore, the function is increasing for x > -2.

In interval notation, this can be written as (-2, ∞).

To summarize:

The function f(x) = 4x + 2 is increasing in the interval (-∞, -2].

The function f(x) = x^2 - 10 is increasing in the interval (-2, ∞).

Thus, the intervals over which the function is increasing, decreasing, or constant are as follows:

Increasing: (-2, ∞)
Decreasing: None (DNE)
Constant: (-∞, -2]

To determine the intervals over which the function is increasing, decreasing, or constant, we need to find the critical points of the function f(x). These are the points where the derivative of f(x) changes sign.

First, let's find the derivative of f(x). Differentiating each part of the piecewise function separately, we have:

For x ≤ -2: f'(x) = d/dx (4x + 2) = 4.

For x > -2: f'(x) = d/dx (x^2 - 10) = 2x.

Next, let's find the critical points by setting the derivative equal to zero and solving for x:

For x ≤ -2: 4 = 0 has no solution.

For x > -2: 2x = 0 gives x = 0.

Now we have the critical point x = 0. We need to determine the intervals between this critical point and the endpoints of the domain x ≤ -2 and x > -2.

For x ≤ -2, the function f(x) = 4x + 2 is a line with a positive slope (4), so it is increasing in this interval. The interval can be expressed as (-∞, -2].

For x > -2, the function f(x) = x^2 - 10 is a quadratic with a positive coefficient for x^2, so it opens upward and is increasing. The interval can be expressed as (0, +∞).

Therefore, the function f(x) is increasing over (-∞, -2] ∪ (0, +∞). It is constant at x = -2 since both parts of the piecewise function are equal at this point.

To summarize:
- The function is increasing over the intervals (-∞, -2] and (0, +∞).
- The function is constant at x = -2.
- The function is undefined at x = 0 since it does not belong to any part of the domain.

even without calculus, you can do these. Just think of what you know about lines and parabolas.

Then you can verify your Algebra I with calculus.

What do you think?