If f={(1,2),(-3,2),(2,5)} and g={(2,4),(1,5),(3,2)}.find:
a)f+g and f-g
b)the domain of (f+g)
the domain of f+g is D(f) ∩ D(g) = {1,2}
For other values, both f and g are not defined, so f+g cannot be defined.
f+g = {(1,2+5),(2,5+5)} = {(1,7),(2,10)}
do f-g the same way.
a) To find f + g, we need to add the corresponding elements of f and g.
f + g = { (1+2, 2+4), (-3+1, 2+5), (2+3, 5+2) }
= { (3, 6), (-2, 7), (5, 7) }
To find f - g, we need to subtract the corresponding elements of g from f.
f - g = { (1-2, 2-4), (-3-1, 2-5), (2-3, 5-2) }
= { (-1, -2), (-4, -3), (-1, 3) }
b) The domain of (f + g) is the set of x-values in (f + g).
Domain of (f + g) = { 3, -2, 5 }
Good
If the intersection of f+g is 1 and 2 son the f+g is 3
To find f + g, we need to add the corresponding elements of f and g.
f = {(1,2), (-3,2), (2,5)}
g = {(2,4), (1,5), (3,2)}
Adding the corresponding elements:
f + g = {(1+2,2+4), (-3+1,2+5), (2+3,5+2)}
= {(3,6), (-2,7), (5,7)}
So, f + g = {(3,6), (-2,7), (5,7)}.
To find f - g, we need to subtract the corresponding elements of g from f.
f - g = {(1-2,2-4), (-3-1,2-5), (2-3,5-2)}
= {(-1,-2), (-4,-3), (-1,3)}
So, f - g = {(-1,-2), (-4,-3), (-1,3)}.
Now, let's find the domain of (f + g).
The domain of a function is the set of all possible x-values or inputs for which the function is defined.
In this case, (f + g) = {(3,6), (-2,7), (5,7)}. The domain will be the set of x-values from these ordered pairs.
Domain of (f + g) = {3, -2, 5}
So, the domain of (f + g) is {3, -2, 5}.