Final answer:
The value of x at which the function f(x) = (3+x)e^{-4x} has a possible relative minimum is x = 2.75, found by setting the first derivative equal to zero. The second derivative test confirms that it is a relative minimum, indicating a stable equilibrium.
Step-by-step explanation:
To find the value of x at which the function f(x) = (3+x)e^{-4x} has a possible relative maximum or minimum point, we need to differentiate the function and find its critical points. First, we take the first derivative of the function with respect to x:
f'(x) = d/dx[(3+x)e^{-4x}] = e^{-4x} - 4(3+x)e^{-4x}
Setting the first derivative equal to zero to find critical points gives us:
e^{-4x}(1 - 4x - 12) = 0
Since e^{-4x} > 0 for all x, we're left with:
1 - 4x - 12 = 0
x = 2.75
Next, we take the second derivative to determine the nature of the critical point:
f''(x) = d^2/dx^2[(3+x)e^{-4x}] = 16(3+x)e^{-4x} - 4e^{-4x}
Substituting x = 2.75 into the second derivative:
f''(2.75) = 16(5.75)e^{-11} - 4e^{-11} > 0
The second derivative is positive at x = 2.75, indicating that this point is a relative minimum and represents a stable equilibrium.