range of xsqrt(36-x) — step-by-step solution

⚠ xsqrt(36-x) is read as x*sqrt(36-x): use * for multiplication, use ^ for a power

1) Let's find the real values taken by

x \cdot \sqrt{36 - x}

using its variation table

⇥Let's study the variations of

f (x)

: first find where its derivative

f^{'} (x) = \sqrt{36 - x} - \frac{x}{2 \cdot \sqrt{36 - x}}

is zero

⇥⇥1.1) Let's solve in

ℝ

\sqrt{36 - x} - \frac{x}{2 \cdot \sqrt{36 - x}} = 0

⇥⇥We multiply both sides by

2 \cdot \sqrt{36 - x}

, nonzero for the fractions to be defined

⇥⇥

\sqrt{36 - x} \cdot 2 \cdot \sqrt{36 - x} - \frac{x}{2 \cdot \sqrt{36 - x}} \cdot 2 \cdot \sqrt{36 - x} = 0 \cdot 2 \cdot \sqrt{36 - x}

⇥⇥ⓘ Multiplying powers with the same base adds the exponents:

\sqrt{36 - x} \cdot \sqrt{36 - x}

36 - x

■Multiplying a fraction by a term multiplies its numerator:

\frac{x}{2 \cdot \sqrt{36 - x}} \cdot 2 \cdot \sqrt{36 - x}

\frac{x \cdot 2 \cdot \sqrt{36 - x}}{2 \cdot \sqrt{36 - x}}

■A product with a zero factor is zero:

0 \cdot 2 \cdot \sqrt{36 - x}

0

⇥⇥

(36 - x) \cdot 2 - \frac{x \cdot 2 \cdot \sqrt{36 - x}}{2 \cdot \sqrt{36 - x}} = 0

⇥⇥ⓘ The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor:

- \frac{x \cdot 2 \cdot \sqrt{36 - x}}{2 \cdot \sqrt{36 - x}}

\frac{2 \cdot (- x)}{2}

⇥⇥

(36 - x) \cdot 2 + \frac{2 \cdot (- x)}{2} = 0

⇥⇥ⓘ The minus sign is placed in front of the product:

2 \cdot (- x)

- 2 x

⇥⇥

(36 - x) \cdot 2 - \frac{2 x}{2} = 0

⇥⇥ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

- \frac{2 x}{2}

- x

⇥⇥

(36 - x) \cdot 2 - x = 0

⇥⇥ⓘ Expand by multiplying each term inside the parentheses:

(36 - x) \cdot 2

72 - x \cdot 2

⇥⇥

72 - x \cdot 2 - x = 0

⇥⇥ⓘ Like terms are grouped by adding their coefficients:

- x \cdot 2 - x

-3 x

⇥⇥ⅰ

\sqrt{36 - x} - \frac{x}{2 \cdot \sqrt{36 - x}}

72 - 3 x

⇥⇥

72 - 3 x = 0

⇥⇥ⓘ We add

3 x

to both sides

⇥⇥

72 = 3 x

⇥⇥ⓘ Both sides are divided by

3

⇥⇥

x = \frac{72}{3}

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

3

, their greatest common divisor:

\frac{72}{3}

24

⇥⇥Candidates, pending the domain check:

x = 24

⇥⇥1.2) Let's evaluate the domain of

\sqrt{36 - x} - \frac{x}{2 \cdot \sqrt{36 - x}} = 0

⇥⇥ For the expression

\sqrt{36 - x}

to be defined, we must have:

36 - x \geq 0

⇥⇥ For the expression

- \frac{x}{2 \cdot \sqrt{36 - x}}

to exist, you must check:

2 \cdot \sqrt{36 - x} \neq 0

⇥⇥1.3) Let's solve in

ℝ

36 - x \geq 0

⇥⇥ⓘ We add

x

to both sides

⇥⇥

36 \geq x

⇥⇥Solution found:

x \leq 36

⇥⇥1.4) Let's solve in

ℝ

2 \cdot \sqrt{36 - x} \neq 0

⇥⇥We simplify the expression by

2

⇥⇥⇥Let's solve in

ℝ

\sqrt{36 - x} \neq 0

⇥⇥⇥

\sqrt{36 - x} \neq 0

is equivalent to:

36 - x > 0

⇥⇥⇥ⓘ We add

x

to both sides

⇥⇥⇥

36 > x

⇥⇥⇥Solution found:

x < 36

⇥⇥Intersecting

x \leq 36

and

x < 36

gives

x < 36

⇥⇥Domain constraints:

x < 36

⇥

f^{'} (x)

is zero at

x = 24

⇥Built from usual functions,

f^{'} (x)

is continuous where it is defined: on each interval where it is never zero, it keeps the sign of its value at any point

⇥

f^{'} (0) = 6

> 0:

f

is increasing on

(-∞, 24)

⇥

f^{'} (25) = - \frac{3 \cdot \sqrt{11}}{22}

< 0:

f

is decreasing on

(24, 36)

⇥Let's calculate the values of

f

at the bounds of the table, or its limits at the infinite or excluded bounds

⇥As

x

tends to

-∞

f (x)

tends to

-∞

⇥

f (24) = 24 \cdot \sqrt{36 - 24} = 48 \cdot \sqrt{3}

⇥

f (36) = 36 \cdot \sqrt{36 - 36} = 0

⇥Let's draw up the variation table:

⇥

$x$	$-∞$		$24$		$36$
$f^{'} (x)$		+	$0$	-	‖
$f (x)$	$-∞$	↗	$48 \cdot \sqrt{3}$	↘	$0$

⇥So

f

is increasing on

(-∞, 24]

and decreasing on

[24, 36]

(-∞, 36]

f

is continuous: it takes every value of

(-∞, 48 \cdot \sqrt{3}]

▶Range:

(-∞, 48 \cdot \sqrt{3}]