F(x)=(x+2)^2 the largest open interval of the domain over which the function is increasing
F' = 2(x+2)
so where is F' > 0 ?
To find the largest open interval over which the function is increasing, we need to determine the points where the derivative of the function is greater than zero.
First, let's find the derivative of the function f(x). The derivative of (x+2)^2 can be computed using the chain rule:
f'(x) = 2(x+2)(1) = 2(x+2)
To find where the function is increasing, we set the derivative greater than 0:
2(x+2) > 0
We can solve this inequality by determining the critical points where the derivative is equal to zero:
2(x+2) = 0
x+2 = 0
x = -2
Now, we can test intervals to the left and right of x = -2 to determine where the derivative is positive:
For x < -2: Choose x = -3 (a value less than -2)
f'(-3) = 2(-3+2) = 2(-1) = -2
Since f'(-3) < 0, the derivative is negative for x < -2.
For x > -2: Choose x = 0 (a value greater than -2)
f'(0) = 2(0+2) = 2(2) = 4
Since f'(0) > 0, the derivative is positive for x > -2.
Therefore, the largest open interval over which the function f(x) = (x+2)^2 is increasing is (-2, ∞).
To find the largest open interval over which the function f(x) = (x + 2)^2 is increasing, we need to analyze the behavior of the function.
A function is considered increasing if its derivative is positive.
First, let's find the derivative of f(x) using the power rule of differentiation:
f'(x) = 2(x + 2) * 1
Simplifying, we get:
f'(x) = 2(x + 2)
To determine when this derivative is positive, we set it greater than zero:
2(x + 2) > 0
Now, solve for x:
x + 2 > 0
x > -2
Therefore, the function f(x) = (x + 2)^2 is increasing for all x greater than -2. Since we're looking for the largest open interval, we exclude the value -2 from the interval.
Hence, the largest open interval of the domain over which the function is increasing is (-2, +∞).