f(x)-x^2-4x+k for domain x>=p
State the smallest value of p for which f is one-one.
To determine the smallest value of p for which f(x) is one-one (or injective), we need to ensure that each value of f(x) corresponds to a unique value of x within the specified domain x ≥ p.
For a function to be one-one, every value of f(x) must be unique. This means that for every two different values of x, their corresponding values of f(x) should also be different.
To explore this, let's begin by finding the derivative of f(x) to determine its critical points. The critical points are the values of x where the rate of change of f(x) changes or where f'(x) equals zero.
Given f(x) = x^2 + 4x + k, let's find its derivative:
f'(x) = 2x + 4
Next, let's set f'(x) equal to zero and solve for x:
2x + 4 = 0
2x = -4
x = -2
The critical point of f(x) is x = -2. This means that at x = -2, the rate of change of f(x) changes or f'(x) equals zero.
To determine if f(x) is strictly increasing or decreasing around x = -2, we can evaluate the sign of f'(x) for different values of x.
If f'(x) > 0 for all x > -2, then f(x) is strictly increasing to the right of x = -2.
If f'(x) < 0 for all x > -2, then f(x) is strictly decreasing to the right of x = -2.
Let's evaluate the sign of f'(x) by substituting different values for x:
For x = -3:
f'(-3) = 2(-3) + 4 = -2 < 0
For x = -1:
f'(-1) = 2(-1) + 4 = 2 > 0
Since f'(x) changes sign between x = -3 and x = -1, we can conclude that f(x) is strictly increasing to the right of x = -2.
For f(x) to be one-one, it should not have any repeated values. Therefore, we need to choose a value of p that ensures f(x) does not have any repeated values for x ≥ p.
Since f(x) is strictly increasing to the right of x = -2, the smallest value of p for which f(x) is one-one is p = -2.
Hence, for the domain x ≥ -2, the smallest value of p for which f(x) is one-one is p = -2.