calculus question please help?

Question

Find a value of "a" such that the funciton f(x)=x^2 e^ax has a critical point at x=2

Answer 1

First you have to take the derivative of f(x), because we know that if the slope of the graph at a given point (derivative) is zero, then it will be horizontal at that point and we'll have a critical point. So...

f(x) = x^2 e^ax
f'(x) = 2x e^ax + ax^2 e^ax [derivative]
0 = (2x + ax^2) e^ax [equate it to zero, to find critical points]
0 = (2(2) + a(2)^2) e^ax [substitute x = 2]
0 = (4 + 4a) [e^ax cannot possibly equal 0]
4a = -4
a = -1

Hope this helps.

Answer 2

What about a = -1?

Answer 3

f(x) = x^2 e^ax
df/dx = ax^2 e^ax + 2x e^ax = (ax^2 + 2x) e^ax

For x = 2:

df/dx = (4a + 4) e^2a = 0
4a + 4 = 0 ==> a = -1

That's your answer.

Answer 4

x=2
a=set all of real numbers
you can use anyvalue of a there and put value of two at the x

Answer 5

Why?