Suppose that the functions S and T are define for all real numbers x as follow.

s(x)=x-3
t(x)=3x+1

Write the expressions for (s-t)(x)and (s.t)(x) and evaluate (s+t(-2)

(s-t) = (x-3)-(3x+1) = -2x-4

(s.t) = (x-3)(3x+1)
(s+t)(-2) = s(-2)+t(-2) = -5 + -5 = -10
or,
(s+t)(x) = (x-3)+(3x+1) = 4x-2
so (s+2)(-2) = 4(-2)-2 = -10

To find the expressions for (s-t)(x) and (s.t)(x), we need to substitute the functions S(x) and T(x) into the given expressions.

1. (s-t)(x)
To find (s-t)(x), we subtract the function T(x) from the function S(x):

(s-t)(x) = S(x) - T(x)

Substituting the given functions:
(s-t)(x) = (x-3) - (3x+1)

Simplifying the expression:
(s-t)(x) = x - 3 - 3x - 1
(s-t)(x) = -2x - 4

So, the expression for (s-t)(x) is -2x - 4.

2. (s.t)(x)
To find (s.t)(x), we multiply the function S(x) by the function T(x):

(s.t)(x) = S(x) * T(x)

Substituting the given functions:
(s.t)(x) = (x-3) * (3x+1)

Expanding the expression:
(s.t)(x) = 3x^2 + x - 9x - 3

Simplifying the expression:
(s.t)(x) = 3x^2 - 8x - 3

So, the expression for (s.t)(x) is 3x^2 - 8x - 3.

3. Evaluating (s+t(-2))
To evaluate (s+t(-2)), we substitute -2 into the function T(x):

T(x) = 3(-2) + 1
T(x) = -6 + 1
T(x) = -5

Then, we substitute the result back into the function S(x) and add it to T(x):

S(x) = x - 3
S(-2) = -2 - 3
S(-2) = -5

(s+t(-2)) = S(-2) + T(x)
(s+t(-2)) = -5 + (-5)
(s+t(-2)) = -5 - 5
(s+t(-2)) = -10

So, (s+t(-2)) equals -10.