domain of f(x)=sqrt((x^2-5x+4)/(3-x))

We save the definition:

f (x) = \sqrt{\frac{x^{2} - 5 x + 4}{3 - x}}

1) Let's find the domain of

\sqrt{\frac{x^{2} - 5 x + 4}{3 - x}}

For the expression

\sqrt{\frac{x^{2} - 5 x + 4}{3 - x}}

to be defined, we must have:

\frac{x^{2} - 5 x + 4}{3 - x} \geq 0

For the expression

\frac{x^{2} - 5 x + 4}{3 - x}

to be defined, we must have:

3 - x \neq 0

2) Let's solve in

ℝ

\frac{x^{2} - 5 x + 4}{3 - x} \geq 0

We study the sign of the numerator

Let's study the sign of

x^{2} - 5 x + 4

by solving in

ℝ

x^{2} - 5 x + 4 \geq 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -5

and

c = 4

Δ = {(-5)}^{2} - 4 \cdot 1 \cdot 4

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

{(-5)}^{2}

(-5) \cdot (-5)

25

Δ = 25 - 4 \cdot 4

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 4

16

Δ = 25 - 16

ⓘ Start from

25

and move back

16

on the number line:

25 - 16

9

Δ = 9

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-5) - \sqrt{9}}{2 \cdot 1}

x_{1} = \frac{5 - \sqrt{9}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{9}

3

;

3^{2}

9

x_{1} = \frac{5 - 3}{2}

ⓘ The numerator and the denominator are computed before the division:

5 - 3

2

x_{1} = \frac{2}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{2}{2}

1

x_{1} = 1

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-5) + \sqrt{9}}{2 \cdot 1}

x_{2} = \frac{5 + \sqrt{9}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{9}

3

;

3^{2}

9

x_{2} = \frac{5 + 3}{2}

ⓘ The numerator and the denominator are computed before the division:

5 + 3

8

x_{2} = \frac{8}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{8}{2}

4

x_{2} = 4

The quadratic has the sign of a =

1

outside its roots, so it is indeed positive outside its roots; the roots, where it is zero, are included.

Solution obtained:

x \in (-∞, 1] \cup [4, +∞)

⇥So the expression

x^{2} - 5 x + 4

is positive or zero on

(-∞, 1] \cup [4, +∞)

and negative otherwise

We study the sign of the denominator, which must not be zero

⇥The expression

3 - x

is zero at:

⇥

x = 3

⇥Its coefficient of

x

-1

, being negative, it is positive before and negative after

3

(forbidden value)

3) Let's draw up the sign table:

x	$-∞$		$1$		$3$		$4$		$+∞$
$x^{2} - 5 x + 4$		+	0	-		-	0	+
$3 - x$		+		+	0	-		-
$\frac{x^{2} - 5 x + 4}{3 - x}$		+	0	-	❌	+	0	-

Solution found:

x \in (-∞, 1] \cup (3, 4]

4) Let's solve in

ℝ

3 - x \neq 0

ⓘ We add

x

to both sides

3 \neq x

Solution obtained:

x \neq 3

Let's build the table of the different terms to determine the global solution:

x	$-∞$		$1$		$3$		$4$		$+∞$
$\frac{x^{2} - 5 x + 4}{3 - x} \geq 0$		✅	✅	❌	❌	✅	✅	❌
$3 - x \neq 0$		✅	✅	✅	❌	✅	✅	✅
Domain of definition:		✅	✅	❌	❌	✅	✅	❌

▶Domain of definition:

x \in (-∞, 1] \cup (3, 4]