Real function in 2 var

Could anyone help me?

here's the equation

f(x,y)= e^sqrt[(x(y^2+1))/(y+2)]+1

1. Compute and draw the domain.

2. Compute and draw the level curve at 2

3. Does the equation admit any global minima?

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You need to notice that the radical function does not exist for negative values, hence, you need to request the following, such that:

`(x(y^2+1))/(y+2) >= 0`

Since `y^2 + 1` is positive all the time, hence, `x` and `y+2` need to be also positive for the entire fraction to rest positive, such that:

`x >= 0`

`y + 2 >= 0 => y >= -2`

**Hence, the domain of the function is the region located above the line `y = -2` and to the right of `x = 0` .**

You need to check if the function has a global minima, hence, you need to perform the second derivative test for two variable functions, such that:

`D(x_c,y_c) = f_(x x)(x_c,y_c)f_(y y)(x_c,y_c) - (f_(x y)(x_c,y_c))^2`

You need to evaluate the partial derivatives `f_x` and `f_y` such that:

`f_x = (e^sqrt((x(y^2+1))/(y+2)))*(((y^2+1)/(y+2))/(2sqrt((x(y^2+1))/(y+2))))`

`f_y = (e^sqrt((x(y^2+1))/(y+2)))*((2xy(y+2) - x(y^2+1))/((y+2)^2))/(2sqrt((x(y^2+1))/(y+2)))`

`f_y = (e^sqrt((x(y^2+1))/(y+2)))*((xy^2 + 4xy - x)/((y+2)^2))/(2sqrt((x(y^2+1))/(y+2)))`

Since `f_x!=0` , hence, you need to check if `f_y = 0` such that:

`x(y^2 + 4y - 1) = 0 => x = 0`

`y^2 + 4y - 1 = 0 => y_(1,2) = (-4+-sqrt(16 + 4))/2 => y_(1,2) = -2+-sqrt5`

You need to evaluate `f_(x x)(0,-2+-sqrt5), f_(y y)(0,-2+-sqrt5) ` and f`_(x y)(0,-2+-sqrt5)` such that:

`f_(x x)= (e^sqrt((x(y^2+1))/(y+2)))*(((y^2+1)/(y+2))^2/(4((x(y^2+1))/(y+2))))` + `(e^sqrt((x(y^2+1))/(y+2)))*(-((y^2+1)/(y+2))^2/(sqrt(x(y^2+1)/(y+2))))/(4(x(y^2+1)/(y+2)))`

**Notice that you cannot evaluate `f_(x x)(0,-2+-sqrt5)` , since it does not exists for `x = 0,` hence, the function does not have global minima.**

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