source("http://macosa.dima.unige.it/r.R") # If I have not already loaded the library
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*S** *02** Inequalities and systems of inequalities**
**diseq**(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, or use **Diseq**.
**diseq1**(F, k, col) punctuates where the fun. of 2 var. F is lower than k
**diseq2**(F, k, col) punctuates where the fun. of 2 var. F is greater than k
**Diseq**, **Diseq1**, **Diseq2** repeat the same action 5 times.
I can also solve** systems of inequalities**
F = 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(G,F,-1,3,"orange")
# or:
Diseq(F,G,-1,3,"black"); Diseq(G,F,-1,3,"orange")
* *
* *
# For the chart on the right ( |x^2-3*x-3| < |x^2+5*x-8| ) see here
* *
k = 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")
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") **# false** (val. < 1)
Diseq2(P,0, "red") **# true** (val. > 0)
Diseq1(P,1,"blue"); 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 graphs **do 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")
# 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")
#
# For **linear programming** see here.
**Other examples of use**