what is the minimum value of f(x)=3xe^x
To find the minimum value of the function f(x) = 3xe^x, we can use calculus.
Step 1: Find the derivative of the function f(x) with respect to x.
The derivative of f(x) is given by:
f'(x) = 3e^x + 3xe^x
Step 2: Set the derivative equal to zero and solve for x.
3e^x + 3xe^x = 0
We can factor out e^x:
e^x(3 + 3x) = 0
Setting each factor equal to zero:
e^x = 0 (this is not possible since e^x is always positive)
3 + 3x = 0
3x = -3
x = -1
Step 3: Determine if it is a minimum or maximum.
To determine if it is a minimum or maximum, we need to find the second derivative.
The second derivative of f(x) is given by:
f''(x) = 3e^x + 3e^x + 3xe^x = 6e^x + 3xe^x
Step 4: Calculate the second derivative at x = -1.
Substituting x = -1 into the second derivative formula:
f''(-1) = 6e^(-1) + 3(-1)e^(-1)
= 6/e - 3/e
= (6-3)/e
= 3/e
Since the second derivative is positive at x = -1 (3/e > 0), this means that the point x = -1 corresponds to a local minimum value of the function f(x).
Step 5: Calculate the value of f(-1).
Substituting x = -1 into the original function f(x):
f(-1) = 3(-1)e^(-1)
= -3/e
Therefore, the minimum value of the function f(x) = 3xe^x occurs at x = -1, and it is equal to -3/e.
As x approaches negative infinity, the value of e^x approaches zero and the value of f(x) approaches zero as well. However, since x cannot be negative infinity and f(x) is increasing for x > 0, the minimum value of f(x) occurs at x = 0. Plugging x = 0 into the function, we get:
f(0) = 3(0)e^0 = 0
Therefore, the minimum value of f(x) is 0.