Question

q 14 fast please whoever will answer will get a thanks ​

Answer 1

Answer:

f : R -> R

f = {(x, x² / 1 - x²) : -∞<x<∞ }

f(x) = x² / 1 - x²

Domain of f :

1 - x² ≠ 0

x² ≠ 1

x ≠ -1 , 1

Thus Domain : x ∈ ℝ - {-1,1}

Range of f :

y = x² / 1 - x²

y (1-x²) = x²

y - yx² = x²

yx² + x² = y

x² = y / 1+y

1+y ≠ 0

y ≠ -1

& y / 1+y ≥ 0

x = f(y) = y / 1+y

let's graph f(y) = y / 1+y

Horizontal line intercept (Y int in this case)

x = 0 implies y = 0 thus (0,0) ∈ f(y)

Asymptotes :

Vertical

When y = -1

Horizontal

When x = 1

With this infornation of Asymptotes and horizontal line intercept

we can graph the function f(y) = y / 1+y

Thus Using the graph f(y) = y / 1+y {I have attached the graph in the answer}

we can easily find when y / 1+ y ≥ 0

y / 1+ y ≥ 0 ∀ y ∈ (-∞, -1) ∪ [0,∞)

Thus Range : y ∋ (-∞,-1) ∪ [0,∞)