source("http://macosa.dima.unige.it/r.R") # If I have not already loaded the library ---------- ---------- ---------- ---------- ---------- ---------- ---------- ----------S02Inequalities and systems of inequalitiesdiseq(F,G, h,k, col) punctuates (using the color col) the space between the graphs of F and G [F(x) < G(x)] and the vertical lines x=h, x=k. If necessary, repeat the command to get additional points.diseq1(F, k, col) punctuates where the fun. of 2 var. F is lower than kdiseq2(F, k, col) punctuates where the fun. of 2 var. F is greater than k I can also solvesystems of inequalitiesF <- function(x) x*(x-2); G <- function(x) x/2 graphF(F,-1,3, "blue"); graph(G,-1,3, "red") # See the charts below to the left diseq(F,G,-1,3,"black") diseq(F,G,-1,3,"black") diseq(G,F,-1,3,"orange") diseq(G,F,-1,3,"orange")# For the chart on the right ( |x^2-3*x-3| < |x^2+5*x-8| ) see herek <- function(x,y) abs(x)+abs(y)-1 # see below on the left PLANE(-2,2, -2,2); CURVE(k,"blue"); diseq1(k,0,"red") # h <- function(x,y) (x-1)^2+0.5*(y+3)^2 # see below in the center # where h < 2, where h > 3 PLANE(-3,3, -6,0) diseq1(h,2, "blue"); diseq1(h,2, "blue") diseq2(h,3, "red"); diseq2(h,3, "red") # P <- function(x,y) x >= y & x-y <=1 & 0 <= x & x <= 1 # where P is true (val. 1), where is false (val. 0) (see) PLANE(-2,2, -2,2) # see below on the right diseq1(P,1,"blue"); diseq1(P,1,"blue")# false(val. < 1) diseq2(P,0, "red"); diseq2(P,0, "red")# true(val. > 0) for(i in 1:10) diseq1(P,1,"blue"); for(i in 1:10) diseq2(P,0,"red")#Caution: the software is not enough # Let's see the resolution of 3/(4*x – 5) > 4. Plane(-5,5, -5,5) f <- function(x) 3/(4*x-5); g <- function(x) 4 graph(f,-5,5, 1); graph(g,-5,5, "red") # I understand that the solutions are in a range between 0 and 2, but to represent the # inequality I need an interval in which the two graphsdo not jump. Where is the jump? # I solve the equation 4*x-5=0. # It's easy, but let's see how to do it with the computer: h <- function(x) 4*x-5; soluz(h,0, 0,2)#1.25 Plane(0,2,-1,8); graph(f,0,2, 1); graph(g,0,2, "red") diseq(g,f, 1.25,2, "brown"); diseq(g,f, 1.25,2, "brown") # I have a graphic confirmation: punctuation shows that the solution is the interval # (1.25,q). To find q I realize that the curves are between 1.25 and 1.5: soluz2(f,g, 1.25,1.5)#1.4375 # To have the interval in a fractional form: fraction(1.25); fraction(1.4375)#5/4 23/16 Point(5/4,0,"blue"); Point(23/16,0,"red") # # Forlinear programmingsee here.Other examples of use