domain and range f(x)=root(-5-6x-x2) - Maths - Relations and Functions

Question

domain and range f(x)=root(-5-6x-x2)????

Gursheen kaur. · Accepted Answer

$f (x) = \sqrt{- 5 - 6 x - x^{2}} D o m a i n o f f, i f \sqrt{- 5 - 6 x - x^{2}} \neq 0 O n s q u a r i n g, w e g e t, - 5 - 6 x - x^{2} \neq 0 o r x^{2} + 6 x + 5 \neq 0 x^{2} + 5 x + x + 5 \neq 0 x (x + 5) + 1 (x + 5) \neq 0 (x + 1) (x + 5) \neq 0 x \neq - 1 a n d x \neq - 5 ∴ D_{f} = s e t o f a l l r e a l n u m b e r s e x c e p t {- 5, - 1}$

Now,

Range of f,

$L e t y = \sqrt{- 5 - 6 x - x^{2}} O n s q u a r i n g, w e g e t, y^{2} = - 5 - 6 x - x^{2} \Rightarrow y^{2} + x^{2} + 6 x + 5 = 0 \Rightarrow y^{2} + 0. y + (x^{2} + 6 x + 5) = 0 y = \frac{0 \pm \sqrt{0 - 4 (x^{2} + 6 x + 5)}}{2} = \frac{\pm \sqrt{- 4 (x^{2} + 6 x + 5)}}{2} = \frac{\pm 2 \sqrt{- (x^{2} + 6 x + 5)}}{2} = \sqrt{- (x^{2} + 6 x + 5)} y = \sqrt{- (x + 5) (x + 1)} ....... (i) T h e l i n e o f s y m m e t r y o f (x^{2} + 6 x + 5) i s, x = \frac{- 6}{2} = - 3 P u t t h e v a l u e o f x = - 3 i n (i), w e g e t, y = \sqrt{- (- 3 + 5) (- 3 + 1)} = \sqrt{- (2) (- 2)} = \sqrt{4} y = 2 ∴ R_{f} = [2, 0]$

domain and range........ f(x)=root(-5-6x-x2)????

domain and range........ f(x)=root(-5-6x-x²)????