Have to find the intervals on which
f(x) = 4 - sqrt(x+2) is increasing. Also the intervals on which the function is decreasing.
TIA
Find the derivative of f(x). When the derivative is positive, the function increases as x increases. When the derivative is negative, the function decreases as x increases.
To find the intervals on which the function f(x) = 4 - sqrt(x+2) is increasing or decreasing, we need to find the derivative of the function.
First, let's find the derivative f'(x) of the function f(x).
f(x) = 4 - sqrt(x+2)
To find the derivative, we can use the power rule and the chain rule.
Power Rule:
If f(x) = x^n, then f'(x) = n*x^(n-1).
Chain Rule:
If f(x) = g(h(x)), then f'(x) = g'(h(x)) * h'(x).
In our case, we have f(x) = 4 - sqrt(x+2). Let's apply the chain rule to find the derivative.
f'(x) = d(4 - sqrt(x+2))/dx
= d(4)/dx - d(sqrt(x+2))/dx
Now, let's compute the derivatives of each term separately.
d(4)/dx = 0, as 4 is a constant.
To compute d(sqrt(x+2))/dx, we can rewrite it as (x+2)^(1/2) and apply the power rule.
d(sqrt(x+2))/dx = (1/2)*(x+2)^(-1/2) * d(x+2)/dx
= (1/2)*(x+2)^(-1/2) * 1
So, d(sqrt(x+2))/dx = (1/2)*(x+2)^(-1/2).
Now, let's substitute these values back into f'(x):
f'(x) = 0 - (1/2)*(x+2)^(-1/2)
= - (x+2)^(-1/2)/2
= -1/(2*sqrt(x+2))
Now that we have the derivative f'(x), we can determine the intervals on which the function f(x) = 4 - sqrt(x+2) is increasing or decreasing.
When f'(x) > 0, the function f(x) is increasing.
When f'(x) < 0, the function f(x) is decreasing.
Let's solve for f'(x) = 0:
-1/(2*sqrt(x+2)) = 0
To find the intervals, we need to determine where the numerator and denominator are zero or undefined.
For the numerator, -1, there are no real solutions as -1 is a constant.
For the denominator, 2*sqrt(x+2) = 0, we solve for x:
2*sqrt(x+2) = 0
sqrt(x+2) = 0
x+2 = 0
x = -2
Therefore, f'(x) is always negative for all values x (except x = -2), which means that the function f(x) = 4 - sqrt(x+2) is decreasing for all real values of x, except at x = -2 (where it is undefined). Hence, f(x) is always decreasing.