f'(x) < 0 if |x| < 1
And apparently this means f is decreasing on (-1, 1).
I don't understand this. I know f'(x) < 0 means the function is decreasing. But I don't understand the absolute value. It means I can put -1 or 1 inside, but wouldn't it still be 1 = 1 and not 1 < 1? I don't understand.
Consider f(x)=x^2 -1
f(x) <0 if abs(x)<1
However, f(x) is not decreasing over that entire range (-1 to 1)
So your meaning of the problem statement is not understood
Sorry, I don't understand your reply...
This question has a prime after the f by the way. It's f prime, aka the derivative. That's why the function is decreasing.
To understand why f'(x) < 0 when |x| < 1 implies that the function f is decreasing on the interval (-1, 1), let's break it down step by step.
First, let's consider the condition |x| < 1. This inequality states that the absolute value of x is less than 1. It means that x can take on any value between -1 and 1, excluding -1 and 1 themselves.
Now, let's focus on the statement f'(x) < 0. This condition says that the derivative of the function f, denoted as f'(x), is always negative when |x| < 1. The derivative measures the rate of change of the function at any given point.
When f'(x) < 0, it means that the function f is decreasing. By definition, a function is decreasing when its values decrease as the input (x) increases.
Now, let's see how these two conditions are related. Since |x| < 1, it implies that x is between -1 and 1. When x is negative, meaning -1 < x < 0, f'(x) < 0 ensures that the function f is decreasing in this interval. Similarly, when x is positive, meaning 0 < x < 1, f'(x) < 0 also guarantees that the function f is decreasing in this interval.
Therefore, by combining the conditions |x| < 1 and f'(x) < 0, we can conclude that the function f is decreasing on the interval (-1, 1).