Math, asked by DangerBoi, 1 year ago

Find the range and domain of the function :
f(x) =  \sqrt{4x -  {(x)}^{2} }

Answers

Answered by ItSdHrUvSiNgH
4

Step-by-step explanation:

Hey user;

firstly;

 y = \sqrt{ 4x - ({x})^{2}} \\ \\ {y}^{2} = 4x - {x}^{2} \\ \\ {y}^{2} - 4x + {x}^{2} \\ \\ D \geq 0 \\ \\ 16 - 4{y}^{2} \geq 0 \\ \\ y \leq 2 \\ \\ Range :- ( - \infty , 2)

 Domain = ( - \infty , \infty)

Answered by abhi569
15

Answer:

Domain is [ 0 , 4 ]

Range is [ 0 , 2 ] .

Step-by-step explanation:

f(x) = √( 4x - x^2 )

For D_{f}, f(x) must be a real number. Since d

f(x) = √( 4x - x^2 ), 4x - x^2 must be a positive real number( including 0 ).

Therefore,

= > 4x - x^2 ≥ 0

= > x( 4 - x ) ≥ 0

= > ( x - 0 )( x - 4 ) ≥ 0

= > x ≥ 4 or x ≤ 0

= > 0 ≥ x ≥ 4

Thus, domain of this function is [ 0 , 4 ].

Let √( 4x - x^2 ) = y

= > 4x - x^2 = y^2

= > x^2 - 4x + y^2 = 0

Since x( domain ) is a real number, discriminant of this must be a real number.

= > Discriminant ≥ 0

= > ( - 4 )^2 - 4( 1 )( y^2 ) ≥ 0

= > 16 - 4y^2 ≥ 0

= > 4 - y^2 ≥ 0

= > ( 2 + y )( 2 - y ) ≥ 0

= > - 2 ≥ y or y ≤ 2

y is the square root of 4x - x^2.

y can't be negative, so range is [ 0 , 2 ] .

Similar questions