Hi
f(x) = g(x) if -x²+3x-2 - ( -x+1) = 0
-x² +3x-2 +x-1 = 0
-x² +4x -3 = 0
To solve, tou have to use the general method of resolution of a quadratic fonction.
To determine if it's has a solution in R, let's calculate Δ
Δ = (4)² - 4 * (1) *(-3)
Δ = 16 +12
Δ= 28
as Δ≥ 0 so the function allow two solution within R
so S 1 = ( -4 +√28) / 2 S 2 = (-4 -√28 ) /2
S1 = ( -4 + 2√7) /2 S2 = (-4 - 2√7) /2
S1 = (2 (-2 +√7) /2 S2 2 (-2 -√7) /2
S1 = -2 +√7 S2 = -2 -√7
So the two function are equal twice. one for x = -2 +√7 and second x = -2-√7
Answer:
The range is { -1,3,7,11,15}
Step-by-step explanation:
The range is just the output values
f(-2) = 4(-2) +7 = -8+7 =-1
f(-1) = 4(-1) +7 = -4+7 = 3
f(0)= 4(0) +7 = 7
f(1) = 4(1) + 7 = 11
f(2) = 4(2) +7 = 8+7 = 15
The answer is 1 and 1/2 feet
Answer:
The relation is not a function
The domain is {1, 2, 3}
The range is {3, 4, 5}
Step-by-step explanation:
A relation of a set of ordered pairs x and y is a function if
- Every x has only one value of y
- x appears once in ordered pairs
<u><em>Examples:</em></u>
- The relation {(1, 2), (-2, 3), (4, 5)} is a function because every x has only one value of y (x = 1 has y = 2, x = -2 has y = 3, x = 4 has y = 5)
- The relation {(1, 2), (-2, 3), (1, 5)} is not a function because one x has two values of y (x = 1 has values of y = 2 and 5)
- The domain is the set of values of x
- The range is the set of values of y
Let us solve the question
∵ The relation = {(1, 3), (2, 3), (3, 4), (2, 5)}
∵ x = 1 has y = 3
∵ x = 2 has y = 3
∵ x = 3 has y = 4
∵ x = 2 has y = 5
→ One x appears twice in the ordered pairs
∵ x = 2 has y = 3 and 5
∴ The relation is not a function because one x has two values of y
∵ The domain is the set of values of x
∴ The domain = {1, 2, 3}
∵ The range is the set of values of y
∴ The range = {3, 4, 5}
