Find the value of a so that the function f(x) = xe^(ax) has critical point at x = 9
I found that the derivative is axe^(ax)+e^(ax)
But I am not sure where to go from here?
e^(ax) (ax+1) = 0 for derivative = 0
so
a = -1/x
a = -1/9
Well, it seems like you've stumbled upon the derivative of the function f(x) = xe^(ax), but now you're stuck in a critical situation. Don't worry, I'm here to clown around and help!
To find the critical point, we know that at that point, the derivative should be equal to zero. So let's set the derivative axe^(ax) + e^(ax) equal to zero:
axe^(ax) + e^(ax) = 0
Now, let's factor out e^(ax):
e^(ax)(ax + 1) = 0
Since e^(ax) can never be equal to zero, we can conclude that (ax + 1) must be equal to zero:
ax + 1 = 0
Now, let's solve for x:
ax = -1
x = -1/a
So, we've found that the critical point occurs at x = -1/a. However, you're specifically looking for the critical point at x = 9. Therefore, we can set x = 9 equal to our solution:
9 = -1/a
To find the value of a, we can rearrange the equation:
a = -1/9
There you have it! The value of a that gives the function f(x) = xe^(ax) a critical point at x = 9 is a = -1/9. Laughter and math, a great combo!
To find the value of "a" that will make the function have a critical point at x = 9, we need to find when the derivative is equal to zero at x = 9.
Let's set up the equation:
f'(x) = axe^(ax) + e^(ax)
We need to determine when f'(x) is equal to zero at x = 9:
0 = 9ae^(9a) + e^(9a)
Now, let's factor out e^(9a) from both terms:
0 = e^(9a)(9a + 1)
For this equation to be true, either e^(9a) = 0 (which is not possible) or (9a + 1) = 0.
Solving for a in (9a + 1) = 0, we get:
9a = -1
a = -1/9
Therefore, the value of "a" that makes the function have a critical point at x = 9 is -1/9.
To find the value of 'a' such that the function f(x) = xe^(ax) has a critical point at x = 9, we need to find the value where the derivative is zero at x = 9.
You correctly found the derivative of f(x) as axe^(ax) + e^(ax). Now we can set this derivative equal to zero and solve for 'a'.
axe^(ax) + e^(ax) = 0
To solve this equation for 'a', we can factor out e^(ax) from both terms:
e^(ax) * (ax + 1) = 0
For the equation to be true, either e^(ax) = 0 (which is not possible since e^x is always positive) OR ax + 1 = 0.
Setting ax + 1 = 0, we can solve for 'a':
ax + 1 = 0
ax = -1
a = -1/x
So, the value of 'a' that gives the function f(x) = xe^(ax) a critical point at x = 9 is a = -1/9.