If g is a differentiable function such that g(x) is less than 0 for all real numbers x and if f'(x)=(x2-4)g(x), which of the following is true?
something is missing
To answer this question, we need to understand the properties of the given function and the relationship between the functions f(x) and g(x). Let's break down the problem step by step.
According to the given information:
1. Function g(x) is differentiable.
2. g(x) is less than 0 for all real numbers x.
3. Function f'(x) is equal to (x^2 - 4)g(x).
Now, let's analyze the options:
a) f(x) is always increasing.
b) f(x) is always decreasing.
c) f(x) has a relative maximum at x = 0.
d) f(x) has a relative minimum at x = 2.
To find the answer, we need to examine the relationship between f'(x) and f(x).
The function f'(x) represents the derivative of f(x), which indicates the rate at which the function f(x) is changing at a particular point.
To determine the nature of f(x), we can look at the sign of f'(x).
Since f'(x) = (x^2 - 4)g(x), we can break it down into two factors:
1. (x^2 - 4): This is a quadratic function that is negative when x is between -2 and 2 and positive otherwise. It has roots at x = -2 and x = 2.
2. g(x): This function is always less than 0, as given in the question.
Multiplying these factors together, we get:
f'(x) = (x^2 - 4)g(x)
From this equation, we can deduce the following:
1. f'(x) is positive when (x^2 - 4) is positive and g(x) is negative.
2. f'(x) is negative when (x^2 - 4) is negative and g(x) is negative.
Now let's analyze each option individually:
a) f(x) is always increasing:
For f(x) to be always increasing, f'(x) should always be positive. However, based on our analysis, f'(x) can be negative for some values of x, which means f(x) cannot be always increasing. Therefore, this option is not true.
b) f(x) is always decreasing:
For f(x) to be always decreasing, f'(x) should always be negative. From our analysis, we know that f'(x) can indeed be negative for some values of x. Additionally, since (x^2 - 4) is negative between -2 and 2, and g(x) is always negative, f'(x) will be negative for all x values outside of -2 to 2. Hence, f(x) will be always decreasing. Therefore, this option is true.
c) f(x) has a relative maximum at x = 0:
To determine if f(x) has a relative maximum at x = 0, we need to examine the sign of f'(x) in the vicinity of x = 0. From our analysis, we know that f'(x) can be both positive and negative, depending on the value of x. Therefore, f(x) does not have a relative maximum at x = 0. Hence, this option is not true.
d) f(x) has a relative minimum at x = 2:
To determine if f(x) has a relative minimum at x = 2, we need to examine the sign of f'(x) in the vicinity of x = 2. From our analysis, we know that f'(x) is positive when (x^2 - 4) is positive and g(x) is negative. Since (x^2 - 4) is positive for x > 2 and g(x) is always negative, f'(x) will be positive for x > 2. Therefore, f(x) does not have a relative minimum at x = 2. Hence, this option is not true.
In conclusion, the only true statement is: b) f(x) is always decreasing.