Specify the region over which the following function f(x) is increasing and the region where it is decreasing. f(x)= √-x
Your answer should be either open intervals or the empty set
ϕ.
To determine the regions where the function f(x) = √(-x) is increasing and decreasing, we need to find the derivative of the function and analyze its sign.
First, let's find the derivative of f(x) with respect to x:
f'(x) = d/dx (√(-x))
Using the chain rule, we can simplify this as follows:
f'(x) = (1/2)(-x)^(-1/2)(-1)
= -(1/2)(√(-x))/(√x)
Now, let's determine where the derivative is positive (increasing) and where it is negative (decreasing).
1. Increasing region:
To find the increasing region, we need to find where f'(x) > 0.
-(1/2)(√(-x))/(√x) > 0
Since the numerator -√(-x) is always negative, our inequality becomes:
√x > 0
As the square root is always positive, this inequality is true for all x > 0.
Therefore, the function f(x) = √(-x) is increasing in the open interval (0, +∞).
2. Decreasing region:
To find the decreasing region, we need to find where f'(x) < 0.
-(1/2)(√(-x))/(√x) < 0
Since the numerator -√(-x) is always negative, our inequality becomes:
√x < 0
However, there are no real values of x for which the square root of a non-positive number is negative. Therefore, the inequality √x < 0 has no solution.
Hence, the function f(x) = √(-x) is neither increasing nor decreasing over any open interval. The answer is the empty set, ϕ.
In summary:
The function f(x) = √(-x) is increasing in the open interval (0, +∞) and is not decreasing over any open interval.