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Real function in 2 varCould anyone help me? here's the equation f(x,y)=...

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raconico | Student, Undergraduate | (Level 1) eNoter

Posted January 3, 2013 at 4:24 PM via web

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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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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted January 4, 2013 at 7:49 AM (Answer #1)

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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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