Find the supremum of the set {x : 3x^2 + 3<10x}?

Answer 1

Find the supremum of the set {x : 3x² + 3<10x}.

3x² + 3 < 10x
3x² -10x + 3 < 0
(3x - 1)(x - 3) < 0
x is an element of (1/3,3).

The supremum is 3.

Answer 2

Hi again :)

3x² + 3<10x
3x² + 3-10x<0
3x² -10x+ 3<0

Notice that the graph of the equation 3x²-10x+3 is a parabole with the form of "U"

The elements that the equation is smaller than 0 is the elements between the two roots of the equation.

Finding the roots:

3x²-10x+3
where:
a=3
b=-10
c=3

delta = b² - 4 *a *c
delta = (-10)² - 4 *3*3 = 100 -36 = 64

x'= (-b+ Vdelta)/2*a
x'= (10+ V64)/2*3
x'= (10+8)/6 = 3

x"= (-b- Vdelta)/2*a
x"= (10- V64)/2*3
x"= (10-8)/6 = 2/6 = 1/3

The roots are {1/3,3} .

So 1/3

Kisses from Brazil

Answer 3

Note that {x : 3x^2 + 3<10x} = (1/3, 3)

The answer follows from the def of sup.

Answer 4

3.