Determine an expression for f(x) in which:

f(x) is cubic,
f(x) ¡Ý 0 when x ¡Ü 2,
f(x) < 0 when x > 2, and
f(0) = 4

I assume you meant:

f(x) >= 0 when x <= 2,
f(x) < 0 when x > 2, and
f(0) = 4

We know that there is a root at x=2, so

f(x) = a(x-2)(bx^2+cx+d)
f(0) = a(-2)(d) = -2ad = 4

That's all we know. So, if we let a=1, d = -2 and

f(x) = (x-2)(bx^2+cx-2)

Unless you have some other info, b and c can be anything.

If I got the definition of f(x) wrong, fix it and follow the logic.

Why did the cubic function go to therapy? Because it couldn't handle all the positive vibes below x = 2 and the negative attitudes above x = 2!

Here's an expression for f(x):

f(x) = (x - 2)(x - 2)(x + 2) + 4

Now, it might look a bit complicated, but let me break it down for you. The expression (x - 2)(x - 2)(x + 2) represents a cubic function that has roots at x = 2 and x = -2. By multiplying three factors together, we ensure that the function is negative for x > 2 and positive for x ≤ 2. And to make f(0) = 4, we add the constant term of 4 at the end.

So, there you have it! The expression for f(x) that satisfies all the given conditions. Keep in mind, this function might have a bit of a split personality due to the positive and negative behavior, but hey, that's what makes it interesting, right?

To determine an expression for f(x) that satisfies the given conditions, we can start by considering a generic cubic function of the form f(x) = ax^3 + bx^2 + cx + d, where a, b, c, and d are constants we need to determine.

Given that f(x) is cubic, we know that the degree of the polynomial is 3.

Since f(0) = 4, we can substitute x = 0 into the expression for f(x) and solve for d:
f(0) = a(0)^3 + b(0)^2 + c(0) + d = d = 4

So, our expression becomes f(x) = ax^3 + bx^2 + cx + 4.

Now let's consider the condition f(x) ≥ 0 when x ≤ 2. This means that the graph of f(x) is above or on the x-axis for x ≤ 2. Since f(0) = 4, the graph of f(x) intersects the y-axis at (0, 4) and is above or on the x-axis when x = 0.

To satisfy this condition, we need the leading coefficient a to be positive, which ensures that the cubic function opens upwards. Let's assume a > 0.

To ensure f(x) ≥ 0 for x ≤ 2, we can set the factors of the cubic function as follows:

1. We want f(2) ≥ 0, so substitute x = 2 into the expression:
f(2) = a(2)^3 + b(2)^2 + c(2) + 4 ≥ 0

2. We want f(0) ≥ 0, which was already satisfied with f(0) = 4.

Since f(0) = 4, we have 4 = 8a + 4b + 2c + 4, which simplifies to 8a + 4b + 2c = 0.

By choosing appropriate values for a, b, and c, we can satisfy the condition f(x) ≤ 0 for x > 2. One way to ensure this is to set the factor (x - 2) in f(x), which will make the numerator positive when x > 2.

Let's finalize our expression for f(x), taking these conditions into account:
f(x) = a(x - 2)(x - k)(x - m) + 4, where k and m are roots of f(x).

By adjusting the values of a, k, and m, we can satisfy the given conditions.

To find an expression for f(x) that satisfies the given conditions, we can start by considering the properties of a cubic function. A general cubic function can be expressed as:

f(x) = ax^3 + bx^2 + cx + d

To ensure that f(x) is cubic, we need the coefficient of x^3 (a) to be non-zero. Let's consider the conditions one by one:

1. f(x) ≥ 0 when x ≤ 2:
To ensure this condition is met, we need the graph of the cubic function to be either above or touching the x-axis when x ≤ 2. This means that the function could have two or three real roots at most. We can consider f(x) = (x - 2)(x - r)^2, where r is a root (a constant) to be determined.

2. f(x) < 0 when x > 2:
To satisfy this condition, we need the function to be negative for all x > 2. This implies that all roots of the function are greater than 2, and the quadratic factor (x - r) is negative for x > 2. To achieve this, we can introduce a negative sign to the quadratic factor:

f(x) = (x - 2)(2 - x)^2

3. f(0) = 4:
We are given that f(0) = 4. Plugging x = 0 into the expression, we get:

4 = (0 - 2)(2 - 0)^2
4 = (-2)(2^2)
4 = (-2)(4)
4 = -8

Since the equation doesn't satisfy the condition, we need to adjust the constant term in the expression. Let's introduce a scaling factor to adjust the function:

f(x) = a(x - 2)(2 - x)^2

Now, we need to determine the value of a such that f(0) = 4. Plugging x = 0 into the expression:

4 = a(0 - 2)(2 - 0)^2
4 = a(-2)(4)
4 = -8a

To solve for a, we divide both sides by -8:

a = -4/8
a = -1/2

Therefore, the expression for f(x) that satisfies the given conditions is:

f(x) = (-1/2)(x - 2)(2 - x)^2