Question

let g be the function given by g(x)=x^2e^kx, where k is a constant. For what value of k does g have a critical point at x=2/3

Answer 1

hello here is a solution 

Answer 2

K= -3

Explanation:

If g(x) be the function given by
`g(x)=x^(2),e^(kx)`

where k is a constant.

We have to find the value of k when the function has a critical point at x =
`(2)/(3)`

Since g'(x) =
`x^(2) e^(kx)`

Now for critical point we will find derivative of g(x) and equate the derivative to zero.

`g'(x)=(d)/(dx)[(x^(2))(e^(kx))]`

g'(x) =
`(2x)(e^(kx))+(kx^(2))(e^(kx))`

`g'(x)=(2x+kx^(2))(e^(kx))`

Now for
`x=(2)/(3)`

`g'((2)/(3))=0`

`[(2((2)/(3))+k((2)/(3))^(2)](e^{((2)/(3)k)})=0`

Sin
`e^(x)` ≠ 0

therefore,
`((4)/(3)+(4)/(9)k)=0`

`(4k)/(9)=-(4)/(3)`

`k=(-4)/(3)((9)/(4))=-3`

Therefore, k = -3 is the answer