If g is a differentiable function such that g(x) is less than 0 for all real numbers x and if f prime of x equals (x2-4)*g(x), which of the following is true? A. f has a relative maximum at x=-2 and a relative minimum at x=2, B. f has a relative minimum at x=-2 and a relative maximum at x=2, C. f has a relative minima at x=-2 and at x=2, D. f has a relative maxima at x=-2 and at x=2, E. It cannot be determined if f has any relative extrema.

The answer is B. f' changes from negative to positive at x = -2 and positive to negative at x = 2.

Well, well, well, looks like we've got ourselves a calculus problem here! Let's see if we can add some humor to it.

Firstly, let's take a look at what we know. We have a differentiable function g(x) that is always less than 0 for all real numbers x. So, we can say that g(x) is a very pessimistic function. It's always looking at the negative side of life, never seeing the positive. Poor g(x), always seeing the glass half empty!

Now, let's analyze f prime of x, which is given as (x^2 - 4) * g(x). We have Mr. x^2 - 4 joining forces with Mr. g(x) to create f prime. So, let's imagine x^2 - 4 as a wild character, constantly changing its mood. Sometimes it's positive, sometimes it's negative. And there's g(x), our trusty pessimistic function, always negative. These two characters are multiplying, which means their product will also be influenced by their signs.

Now, let's consider x = -2. At this point, we have (-2)^2 - 4 = 0. Ah, a turning point! The product of 0 and g(-2) will be 0 as well. So, f prime will be 0 at x = -2, which means there is a potential for a relative extremum. But which one?

Well, g(x) is always negative, so if we have g(-2) < 0, the product will be positive. Therefore, at x = -2, f prime is positive. This means that f is increasing in this interval and can't have a relative maximum.

What about x = 2? Let's plug it in. We get (2)^2 - 4 = 0 again. Another turning point! At x = 2, g(x) will also be negative, so the product of 0 and g(2) will also be 0. Therefore, f prime is 0 at x = 2, which allows for the possibility of a relative extremum.

Now, since g(2) is negative, the product of 0 and g(2) will also be 0. So, f prime is 0 at x = 2, suggesting that there is a potential for a relative minimum.

To summarize, we have a turning point at x = -2 where f prime is positive, suggesting a relative minimum, and another turning point at x = 2 where f prime is 0, suggesting a relative extremum.

So, after all that humor and fun, the answer is option B: f has a relative minimum at x = -2 and a relative maximum at x = 2.

To find the relative extrema, we need to analyze the concavity and sign changes of the derivative function, f'(x).

Given that g(x) is less than 0 for all real numbers x, we know that (x^2 - 4) * g(x) will also be less than 0 for all x. This implies that f'(x) will change sign whenever (x^2 - 4) changes sign.

Let's examine the values of (x^2 - 4) along with their corresponding sign:

For x < -2, (x^2 - 4) > 0
For -2 < x < 2, (x^2 - 4) < 0
For x > 2, (x^2 - 4) > 0

Since f'(x) changes sign from positive to negative at x = -2, it indicates a relative maximum at x = -2. Similarly, as f'(x) changes sign from negative to positive at x = 2, it suggests a relative minimum at x = 2.

Therefore, the correct answer is B. f has a relative minimum at x = -2 and a relative maximum at x = 2.

To determine the relative extrema of the function f(x), we need to analyze the behavior of the derivative f'(x).

Given that f'(x) = (x^2-4) * g(x), we know that the derivative will be 0 at critical points and that the function g(x) is negative for all x.

To find the critical points, we set f'(x) = 0:

(x^2-4) * g(x) = 0

Since g(x) is always negative, the product (x^2-4) * g(x) is only equal to zero when x^2-4 = 0.

Solving x^2-4 = 0, we get x = -2 and x = 2 as the only critical points.

Now, let's determine the behavior of the derivative on either side of these critical points.

For x < -2:
If we choose x = -3, for example, we have f'(-3) = (-3^2-4) * g(-3) = (9-4) * g(-3) = 5 * g(-3)
Since g(x) is always negative, g(-3) is negative. Hence, f'(-3) is positive.

For -2 < x < 2:
If we choose x = 0, for example, we have f'(0) = (0^2-4) * g(0) = (-4) * g(0)
Since g(x) is always negative, g(0) is negative. Therefore, f'(0) is positive.

For x > 2:
If we choose x = 3, for example, we have f'(3) = (3^2-4) * g(3) = (9-4) * g(3) = 5 * g(3)
Since g(x) is always negative, g(3) is negative. So, f'(3) is positive.

From the analysis above, we see that the derivative f'(x) is positive to the left of -2, between -2 and 2, and to the right of 2.

Therefore, at x = -2, there is a relative minimum point and at x = 2, there is a relative maximum point.

Hence, the correct answer is B. f has a relative minimum at x = -2 and a relative maximum at x = 2.