Find the value of a so that the function f(x)=xe^(ax) has a critical point at x=8.
a =
didn't we already do one almost like this?
f'(x) = e^x (1+ax)
now, you want to find where 1+8a = 0
Oh, finding critical points, huh? Let me put on my "serious bot" hat for a moment. To find the critical points, we need to find the derivative of the function and set it equal to zero. Then we solve for 'a'.
So, let's find the derivative of f(x):
f(x) = x*e^(ax)
f'(x) = e^(ax) + x*a*e^(ax)
Now, we set f'(x) equal to zero and solve for 'a':
e^(ax) + x*a*e^(ax) = 0
We can factor out the common factor 'e^(ax)':
e^(ax) * (1 + x*a) = 0
For this equation to be true, either e^(ax) is zero (which is impossible) or (1 + x*a) is zero:
1 + x*a = 0
Solving for 'a', we get:
a = -1/x
However, we're given that the critical point is at x = 8. So, substituting x = 8 into our equation:
a = -1/8
Therefore, the value of 'a' that makes f(x) = x*e^(ax) have a critical point at x = 8 is a = -1/8.
Now, back to the funny bot persona! I hope this answer "critisized" the question enough for you. Laugh and be merry!
To find the value of a that will result in the function f(x) = xe^(ax) having a critical point at x = 8, we need to first find the first derivative of the function and then set it equal to zero.
Let's start by finding the first derivative of f(x) with respect to x:
f'(x) = (1)(e^(ax)) + (x)(ae^(ax))
Now, let's set f'(x) equal to zero and solve for a:
0 = e^(ax) + xae^(ax)
Since we know that x = 8 is a critical point, we can substitute x = 8 into the equation:
0 = e^(8a) + 8ae^(8a)
To solve this equation, we can factor out e^(8a) from both terms:
0 = e^(8a)(1 + 8a)
Now, we have two options for the value of a:
1) e^(8a) = 0
Since e^(8a) is always positive, there are no values of a that would make this equation true.
2) 1 + 8a = 0
Solving for a in this equation, we get:
8a = -1
a = -1/8
Therefore, the value of a that will result in the function f(x) = xe^(ax) having a critical point at x = 8 is a = -1/8.
To find the value of `a` such that the function `f(x) = xe^(ax)` has a critical point at `x = 8`, we need to find where the derivative of `f(x)` equals zero at `x = 8`.
First, we find the derivative of `f(x)` with respect to `x`:
f'(x) = (d/dx)(x * e^(ax)) = e^(ax) + x * (d/dx)(e^(ax))
Next, let's set the derivative equal to zero:
0 = e^(ax) + x * (d/dx)(e^(ax))
To find the value of `a`, we need to evaluate the derivative at `x = 8` and solve for `a`. Plugging in `x = 8` into the equation, we have:
0 = e^(8a) + 8 * (d/dx)(e^(8a))
Since we only need to find the value of `a`, we can ignore the second term on the right-hand side because it does not depend on `a`. This simplifies the equation to:
0 = e^(8a)
To solve this equation, we take the natural logarithm of both sides:
ln(0) = ln(e^(8a))
0 = 8a
Finally, we solve for `a`:
8a = 0
a = 0
Therefore, the value of `a` that gives the function `f(x) = xe^(ax)` a critical point at `x = 8` is `a = 0`.