The quadratic mean of two real numbers x and y equals p (x 2 y 2)/2. By computing the arithmetic and quadratic means of different pairs of positive real numbers, formulate a conjecture about their relative sizes and prove your conjecture.?

The quadratic mean of 2 real positive numbers is greater than or equal to the arithmetic mean.

Step-by-step explanation:

x and y      Quadratic Mean       Arithmetic mean

3 and 3                    3                                    3

2 and 3                   2.55                               2.5

3 and  6                  4.74                                4.5

2 and 5                   3.8                                  3.5

2 and 17                 12.1                                   9.5

18 and 28              23.5                                  23

10 and  48             34.7                                  29

The quadratic mean is always greater than the arithmetic mean except when  x and y are the same.

When the difference between the pairs is  small the difference in the means is also small. As that difference increases the difference in the means also increases.

So we conjecture that the quadratic mean is always greater than or equal to the arithmetic mean.

Proof.

Suppose it is true then:

√(x^2 + y^2) / 2) ≥ (x + y)/2       Squaring  both sides:

(x ^2 + y^2) / 2 ≥ (x + y)^2 / 4    Multiply through by 4:

2x^2 +2y^2  ≥  (x + y)^2

2x^2 +2y^2 >=  x^2 + 2xy + y^2

x^2 + y^2 >= 2xy.

x^2 - 2xy + y^2 ≥ 0

(x - y)^2 ≥ 0

This is true  because the square of any real number is positive so the original inequality must also be true.

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3