Alright, we're looking for the critical numbers of the function f(x)=x^5 e^-8x, which are the points where the derivative of the function is equal to zero. Let's go through it step by step:
Step 1: First, note that a critical number of a function is a number in the domain of the function where either the derivative is zero or the derivative doesn't exist. Since we're not dealing with the trigonometric or fraction function here, we only need to find where the derivative is zero.
Step 2: To find the critical numbers, we first have to take the derivative of our function. We're going to use the product rule for derivatives, which states that the derivative of two multiplied functions is the first function times the derivative of the second plus the second function times the derivative of the first.
Step 3: For our function f(x) = x^5 * e^-8x we get:
f'(x)=[5x^4 * e^-8x] + [x^5 * -8e^-8x] =5x^4 * e^-8x - 8x^5 * e^-8x.
Step 4: After calculating the derivative, we need to set the derivative equal to zero and solve for x, as by definition the derivative of a function at a critical point is equal to zero.
Step 5: So we have 0 =5x^4 * e^-8x - 8x^5 * e^-8x. We can see that both terms feature the common factor e^-8x, so let's factorize: 0 = e^-8x * (5x^4 - 8x^5).
Step 6: e^-8x will never be zero, so the expression within the brackets must be zero. So, 5x^4 - 8x^5 = 0, which simplifies to 0 = x^4 * (5 - 8x).
Step 7: Setting this simplified equation to zero tells us that either x^4 = 0 or 5 - 8x = 0, which respectively yield the solutions x = 0 or x = 5/8.
Therefore, the critical numbers for the function f(x) = x^5 e^-8x are 0 and 5/8.