Find the critical points of f(x)=x^4e^-x.
You would use the product rule.
Then what?
set the derivative equal to zero and solve for x
plug those x values back into the original equation to find the critical points
Would the derivative be e^-x (x^4-4x^3)?
almost.
d/dx(e^-x) = -e^-x
so,
df/dx = (4x^3)(e^-x) + (x^4)(-e^-x)
= e^-x (4x^3 - x^4)
= x^3 e^-x (4-x)
set that to zero, which occurs at x=4
luckily, in this case, you would still have gotten the correct answer, since the only mistake was the sign of df/dx.
dy/dx= x^4(e^-x)(-1) + e^-x (4x^3)
= e^-x( -x^4 - 4x^3)
(looks like you forgot the derivative of the exponent -x )
to solve:
e^-x = 0 or -x^4 - 4x^3 = 0
e^-x = 0 has no solution
x^4 + 4x^3 = 0
x^3(x+4) = 0
x = 0 or x = -4
now sub back in
Sub 0 and -4 into f(x). or f'(x)?
f(x)=x^4e^-x
f(0)=0^4e^0 = 0
f(-4)=-4^4e^4 = 13977.13
I feel like i didn't do that right. It seems like a weird number for a critcal point.
must be because the value for x is 4, not -4
f(4)= 4^4 e^-4 = 4.6888
After applying the product rule to the function f(x) = x^4 * e^(-x), we would differentiate each term separately.
First, differentiating x^4 using the power rule, we get:
f'(x) = 4x^3 * e^(-x)
Next, differentiating e^(-x) using the chain rule, we get:
f'(x) = x^4 * (-e^(-x))
Now, we can find the critical points by setting the derivative equal to zero, f'(x) = 0, and solving for x:
0 = 4x^3 * e^(-x) - x^4 * e^(-x)
To simplify the equation, we can factor out e^(-x):
0 = e^(-x) * (4x^3 - x^4)
For the equation to equal zero, either e^(-x) = 0 (which is not possible) or (4x^3 - x^4) = 0.
Therefore, we set (4x^3 - x^4) = 0:
x^3(4 - x) = 0
From this equation, we can see that there are two possible solutions:
1. x^3 = 0, which means x = 0.
2. (4 - x) = 0, which means x = 4.
Therefore, the critical points of the function f(x) = x^4 * e^(-x) are x = 0 and x = 4.