Find all critical points of the function.
f(x) = xe^3x
so x= ?
Show steps or tell me how I find them?
A critical point is an interior point of an interval of the function at which f'(x)=0 or undefined.
If the interval is not given, it is usually understood to be (-∞, ∞).
f(x)=xe3x
f'(x)=xe3x + e3x
f'(x) is therefore defined within the interval (-∞, ∞).
Now find x for which f'(x)=0
f'(x)=0 when x=-(1/3) or e3x=0. The latter condition requires x to be outside the domain (x=-∞).
Therefore the only critical point is x=-(1/3).
To find the critical points of a function, we need to first find the derivative of the function and set it equal to zero. So let's differentiate the function f(x) = xe^3x with respect to x.
Step 1: Find the derivative of f(x) with respect to x using the product rule
f'(x) = (1 * e^3x) + (x * 3e^3x)
= e^3x + 3xe^3x
Step 2: Set the derivative equal to zero and solve for x
e^3x + 3xe^3x = 0
We can see that e^3x is always positive for any real value of x. Therefore, for the sum of two positive terms to equal zero, the second term 3xe^3x must be equal to zero.
Step 3: Set 3xe^3x = 0 and solve for x
3xe^3x = 0
To find the critical points, we need to find the values of x that satisfy this equation. There are two possibilities:
1) Setting x = 0, we have 3(0)e^(3(0)) = 0, which is true.
2) The term e^3x can never be equal to zero since it is always positive. Therefore, the equation 3xe^3x = 0 is only satisfied when x = 0.
Therefore, the critical point for the function f(x) = xe^3x is x = 0.
To find the critical points of the function f(x) = xe^(3x), we need to determine the values of x at which the derivative of the function is either zero or undefined.
Step 1: Take the derivative of the function.
f'(x) = d/dx (xe^(3x))
To take the derivative, we can use the product rule:
f'(x) = (1)(e^(3x)) + (x)(d/dx(e^(3x)))
The derivative of e^(3x) is simply 3e^(3x), so the second term becomes:
f'(x) = e^(3x) + 3xe^(3x)
Step 2: Set the derivative equal to zero and solve for x.
e^(3x) + 3xe^(3x) = 0
To simplify, factor out e^(3x):
e^(3x)(1 + 3x) = 0
Now, set each factor equal to zero:
e^(3x) = 0 or 1 + 3x = 0
The first equation, e^(3x) = 0, has no solution since e^(3x) is always positive.
Solving the second equation, 1 + 3x = 0, we obtain:
3x = -1
x = -1/3
So, the critical point of the function f(x) = xe^(3x) is x = -1/3.