Question

find domain and range of f(X) (x²-x+1 ) ÷ ( x²+X+1)​

Answer 1

y = x2−x+1x2+x+1
Domain is x ∈R
For Range of function:

yx​2 + xy + y = x2 - x + 1
(y-1)x2 + (y+1)x + y-1 = 0

if y = -1
-2x2 -1 -1 = 0
-2x2 -2 = 0
Therefore, y cannot be equal to -1 since x will be a complex value

Now, y not equal to -1.
(y-1)x2 + (y+1)x + y-1 = 0 has real roots

hence, (y+1)2 - 4 (y-1)(y-1) ≥0
y2 + 2y + 1 - 4y2 + 8y - 4 ≥0
-3y2 + 10y - 3 ≥ 0
3y2 - 10y + 3 ≤0
3y2 - 9y - y + 3 ≤ 0
3y (y-3) - 1 (y-3) ≤0
(3y-1)(y-3) ≤ 0
y ∈[13,3]
Hence, range of function is [13,3]

Answer 2

Domain : x ∈R

Domain : x ∈RRange : y ∈[13,3]

Given:

• f(X) = (x²-x+1 ) ÷ ( x²+X+1)

To Find:

• the range and the domain of f(x).

Solution:

y = x2−x+1x2+x+1

Domain is x ∈R

For the Range of function:

yx2 + xy + y = x2 - x + 1

(y-1)x2 + (y+1)x + y-1 = 0

if y = -1

-2x2 -1 -1 = 0

-2x2 -2 = 0

Since, y can't be equal to the -1 since x will be the complex value.

Now,

we know, y not equal to -1.

(y-1)x2 + (y+1)x + y-1 = 0 it has real the roots.

Thus,

(y+1)2 - 4 (y-1)(y-1) ≥0

y2 + 2y + 1 - 4y2 + 8y - 4 ≥0

-3y2 + 10y - 3 ≥ 0

3y2 - 10y + 3 ≤0

3y2 - 9y - y + 3 ≤ 0

3y (y-3) - 1 (y-3) ≤0

(3y-1)(y-3) ≤ 0

y ∈[13,3]

therefore,

the Domain is x ∈R & Range is y ∈[13,3] .

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