Question

find the domain and range of the function f(x)=2+
$f (x) = 2 + \sqrt{x - 1} find \: the \: domain \: and \: range \: of \: the \: function$
​

Answer 1

Answer:

Domain:

x

≤

1

and

x

≥

2

or

x

∣

(

−

∞

,

1

]

∪

[

2

,

∞

)

Range:

y

≥

0

or

y

∣

[

0

,

∞

)

Explanation:

y

=

√

x

2

−

3

x

+

2

=

√

(

x

−

1

)

(

x

−

2

)

;

Domain: under

root should be

≥

0

∴

(

x

−

1

)

(

x

−

2

)

≥

0

When

1

<

x

<

2

sign of

y

is

(

+

)

(

−

)

=

(

−

)

∴

<

0

Therefore for

1

<

x

<

2

;

y

is undefined .

Domain:

x

≤

1

and

x

≥

2

or

x

∣

(

−

∞

,

1

]

∪

[

2

,

∞

)

Range:

y

≥

0

or

y

∣

[

0

,

∞

)

since square root of positive quantity

is also positive.

graph{(x^2-3x+2)^0.5 [-10, 10, -5, 5]}