Part 1;
(0, 0)
Part 2;
(0, 2.5)
Step-by-step explanation:
Part 1
The given system of inequalities is presented as follows;
f(x) < x + 4; f(x) > -x - 3; and f(x) < 5
We check each of the points as follows;
The point (0, 0) in the inequality f(x) < x + 4, gives;
f(0) < 0 + 4
f(0) = 0 < (is less than) 0 + 4 = 4
Therefore, (0, 0) is a solution of the inequality f(x) < x + 4
The point (0, 0) in the inequality f(x) > -x - 3, gives;
f(0) < -0 - 3
f(0) = 0 > (is larger than) -0 - 3 = -3
Therefore, (0, 0) is a solution of the inequality f(x) > -x - 3
The point (0, 0) in the inequality f(x) < 5, gives;
f(0) < 5
f(0) = 0 < (is less than) 5
Therefore, (0, 0) is a solution of the inequality f(x) < 5
The point (-6, 0) in the inequality f(x) > -x - 3, gives;
f(-6) = -6 - 3 = 3
The point (-6, 0) with y = 0 < (is less than) f(-6) = 3, therefore (-6, 0) is not a solution to (not included in the graph of) the inequality f(x) > -x - 3 and therefore to the system of inequalities because at x = -6, the values of f(x) > -x - 3 are larger than 3
The point (-3, 4) in the inequality f(x) < x + 4, gives;
f(-3) = -3 + 4 = 1
The point (-3, 4) with y = 4 < (is larger than) f(-2) = 1, therefore (-3, 4) is not a solution to (not included in the graph of) the inequality f(x) < x + 4 and therefore to the system of inequalities because at x = -3, the values of f(x) < x + 4 are less than 1
The point (4, 6) is not a solution to (not included in the graph of) the inequality f(x) < 5 and therefore to the system of inequalities because f(x) is larger than 5 for all x
Therefore, the point which is part of the solution set of the system of inequalities is (0, 0)
Part 2
The given system of inequalities are, f(x) ≥ 2·x + 2; f(x) ≤ -4·x + 3; and f(x) ≤ 6·x + 5
Plotting the given system of inequalities on MS Excel the point which is part of the solution is given by points which are within the triangular intersection area of the three inequalities, with coordinates, (1/6, 7/3), (-3/4, 1/2), and (-1/5, 19/5)
Therefore, the points, (0, 0), (-2.5, 0) are not solutions because, the y-value of the solution area are all higher than the line y = 0
The point (0. 6.5) is not a solution because the points in the triangular solution area all have x-values lesser than x = 6.5
The point which is part of the solution by examination is the point (0, 2.5) which is a point between the lines y = 19/5 = 3.8, y = 1/2, x = -3/4, and x = 1/6.
