Question

Is `f(x)=(x^2e^x)/(x+2)` increasing or decreasing at `x=-1`?

Answer 1

Thus, `f(x)` is decreasing at `x = -1`.

Explanation:

To determine if `f(x)` is increasing or decreasing at a specific `x`, first of all, you need to compute the derivative.

Here, you can use the quotient rule:

if `f(x) = (g(x))/(h(x))`, then the derivative can be computed as follows:

`f'(x) = (g'(x) h(x) - g(x) h'(x)) /(h^2(x))`

For you, `g(x) = x^2e^x` and `h(x) = x+2`.

The derivatives of `g(x)` and `h(x)` are:

`g'(x) = x^2e^x + 2xe^x` (with the product rule)

`h'(x) = 1`

Thus, your derivative is:

`f'(x) = ((x^2e^x + 2xe^x)(x+2) - x^2e^x * 1) / (x+2)^2`

`= (e^x(x^2 + 2x)(x+2) - e^x x^2) / (x+2)^2`

`= (e^x[(x^2 + 2x)(x+2) - x^2]) / (x+2)^2`

`= (e^x(x^3 - 3x^2 + 4x)) / (x+2)^2`

Now, let's evaluate the derivative at `x = -1`:

`f'(-1) = (e^(-1)((-1)^3 - 3(-1)^2 +4*(-1))) / (-1+2)^2`

`= e^(-1)(-1 - 3 -4)`

`= -8 e^(-1)`

As `e^(-1) > 0`,

`f'(-1) = -8 e^(-1) < 0`.

Thus, `f(x)` is decreasing at `x = -1`.