Name all the values of x that are not in the domain of f(x)=2-x^2/x+5.

Answer: x= -5

2)Find the minimum value of f(x,y)=2x-y+2 for the polygonal convex set determined by this system of inequalities: x >/= 1, x </= 3, y</=0, 1/2x+y</=5

Answer: -0.5

To find the values of x that are not in the domain of f(x) = (2 - x^2) / (x + 5), we need to consider the values that would make the denominator equal to zero. The denominator, x + 5, cannot be zero since dividing by zero is undefined. Therefore, we set x + 5 = 0 and solve for x:

x + 5 = 0
x = -5

Hence, the only value of x that is not in the domain of f(x) is x = -5.

To find the minimum value of f(x, y) = 2x - y + 2 for the polygonal convex set determined by the system of inequalities, we need to find the critical points or vertices of the polygon formed by the inequalities.

1) The inequality x ≥ 1 represents a vertical line passing through x = 1 and extending infinitely to the right.
2) The inequality x ≤ 3 represents a vertical line passing through x = 3 and extending infinitely to the left.
3) The inequality y ≤ 0 represents a horizontal line passing through y = 0 and extending infinitely downwards.
4) The inequality (1/2)x + y ≤ 5 represents a line in slope-intercept form, y = -1/2x + 5, whose slope is -1/2 and y-intercept is 5.

To find the points of intersection, we can solve the various pairs of inequalities. By doing so, we find two points:
- (1,0) is the intersection of x ≥ 1 and y ≤ 0.
- (3,2) is the intersection of x ≤ 3 and (1/2)x + y ≤ 5.

The polygonal convex set is formed by the line segments connecting these points:
- (1,0) to (3,0)
- (3,0) to (3,2)
- (3,2) to (1,0)

Now we can evaluate f(x, y) = 2x - y + 2 at the vertices of this polygon to find the minimum value. These vertices are:
- (1,0)
- (3,0)
- (3,2)

Plugging in the coordinates into f(x, y):
- f(1,0) = 2(1) - 0 + 2 = 4
- f(3,0) = 2(3) - 0 + 2 = 8
- f(3,2) = 2(3) - 2 + 2 = 6

Therefore, the minimum value of f(x, y) = 2x - y + 2 for the given polygonal convex set is -0.5, which occurs at the point (3,2).