Question

Find the value of x at which the function has a possible relative maximum or minimum point. (Recall that ex is positive for all x) Use the second derivative to determine the nature of the function at these points.

f(x)=(3+x)e-4x
What is the value of x at which the function has a possible relative maximum or minimum point?

Answer 1

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.