Find the critical numbers of the function.
f(x) = x^4*e^-8x
Critical point:
A critical point at the interior of the domain of a function is the point where the derivative is zero or undefined.
The domain of the given function
f(x) = x^4*e^-8x
is [-∞∞], so you will need to find the critical points on that interval.
Since there are no discontinuities, nor vertical asymptotes, you only have to worry about the extrema where f'(x)=0.
Calculate f'(x), equate to zero and solve for x in f'(x)=0. These are the critical points.
Hint: I only find one critical point, and it is quite obvious.
I found one critical point, but there are two. The critical point I found is 0, but I can't find the second one.
Yes, you're right, there are two critical points, namely two points at which f'(x)=0.
I erred in the calculations because I took the wrong function to start with.
The derivative is:
f'(x)=-4x³(2*x-1)e-8*x
which obviously has two roots. I am quite sure you would be able to spot both roots.
Why did the function go to therapy? Because it had too many critical issues! But don't worry, I'm here to help. Let's find those critical numbers for f(x) = x^4*e^-8x.
To find the critical numbers, we need to find where the derivative of the function is equal to zero or undefined. So, let's first find the derivative of f(x).
f'(x) = d/dx (x^4*e^-8x)
We can use the product rule here: (uv)' = u'v + uv'
u = x^4
v = e^-8x
Let's find the derivative of u and v separately:
u' = d/dx (x^4) = 4x^3
v' = d/dx (e^-8x) = -8e^-8x
Now, let's use the product rule:
f'(x) = u'v + uv'
= (4x^3)*(e^-8x) + (x^4)*(-8e^-8x)
= 4x^3*e^-8x - 8x^4*e^-8x
To find the critical numbers, we need to set f'(x) equal to zero and solve for x:
4x^3*e^-8x - 8x^4*e^-8x = 0
We can factor out an e^-8x:
e^-8x * (4x^3 - 8x^4) = 0
Now, we have two factors: e^-8x = 0 and 4x^3 - 8x^4 = 0.
Since e^-8x is never equal to zero for any value of x, we can ignore that factor.
Now, let's solve 4x^3 - 8x^4 = 0:
4x^3 - 8x^4 = 0
4x^3(1 - 2x) = 0
So, either 4x^3 = 0 or 1 - 2x = 0.
For 4x^3 = 0, we get x = 0.
For 1 - 2x = 0, we get x = 1/2.
So, the critical numbers of the function f(x) = x^4*e^-8x are x = 0 and x = 1/2. Keep in mind that these are potential critical points, and further analysis may be needed to determine their nature. Happy crunching!
To find the critical numbers of a function, we need to first find the derivative of the function and then solve for values of x that make the derivative equal to zero or undefined.
The derivative of the given function f(x) = x^4 * e^(-8x) can be found using the product rule and chain rule of differentiation. Let's do that:
1. Apply the product rule: (u * v)' = u' * v + u * v'
For the function f(x) = x^4 * e^(-8x), let's split it into two parts:
u = x^4
v = e^(-8x)
Now, let's find the derivatives of u and v:
u' = 4x^3 (using the power rule of differentiation)
v' = -8e^(-8x) (using the chain rule and derivative of e^u)
2. Apply the product rule: (u * v)' = u' * v + u * v'
f'(x) = u' * v + u * v'
= (4x^3) * (e^(-8x)) + (x^4) * (-8e^(-8x))
= 4x^3e^(-8x) - 8x^4e^(-8x)
Now, to find the critical numbers, we need to solve for values of x that make the derivative f'(x) equal to zero.
Setting f'(x) = 4x^3e^(-8x) - 8x^4e^(-8x) = 0, we can factor out e^(-8x):
4x^3e^(-8x) - 8x^4e^(-8x) = 0
e^(-8x)(4x^3 - 8x^4) = 0
e^(-8x) is never zero, so we can ignore it for finding the critical numbers.
Now, solve the equation 4x^3 - 8x^4 = 0. Let's factor out common terms:
4x^3(1 - 2x) = 0
To find the critical numbers, we set each factor equal to zero:
1 - 2x = 0
2x = 1
x = 1/2
Therefore, the critical number of the function f(x) = x^4 * e^(-8x) is x = 1/2.