range of (3x+4)/(x^2-25) — step-by-step solution

1) Let's find the real values taken by

\frac{3 x + 4}{x^{2} - 25}

using its variation table

⇥Let's study the variations of

f (x)

: first find where its derivative

f^{'} (x) = - \frac{3 x^{2} + 8 x + 75}{{(x^{2} - 25)}^{2}}

is zero

⇥Let's solve in

ℝ

- \frac{3 x^{2} + 8 x + 75}{{(x^{2} - 25)}^{2}} = 0

⇥A quotient is zero if and only if its numerator is zero and its denominator is not: we solve the numerator equal to 0

⇥⇥1.1) Let's solve in

ℝ

- (3 x^{2} + 8 x + 75) = 0

⇥⇥ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (3 x^{2} + 8 x + 75)

- 3 x^{2} - 8 x - 75

⇥⇥

- 3 x^{2} - 8 x - 75

⇥⇥

- 3 x^{2} - 8 x - 75 = 0

⇥⇥Since

-1 \neq 0

, divide both sides by

-1

3 x^{2} + 8 x + 75 = 0

⇥⇥Given the quadratic form

a x^{2} + b x + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 3

b = 8

and

c = 75

⇥⇥

Δ = 8^{2} - 4 \cdot 3 \cdot 75

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

8^{2}

8 \cdot 8

64

■Multiplications are done before additions and subtractions:

4 \cdot 3

12

⇥⇥

Δ = 64 - 12 \cdot 75

⇥⇥ⓘ Multiplications are done before additions and subtractions:

12 \cdot 75

(10 + 2) \cdot 75

10 \cdot 75 + 2 \cdot 75

750 + 150

900

⇥⇥

Δ = 64 - 900

⇥⇥ⓘ Start from

64

and move back

900

on the number line, going below zero:

64 - 900

-836

⇥⇥

Δ = -836

⇥⇥As

Δ < 0

, the equation has no real root.

⇥⇥Solution:

x \in \emptyset

⇥⇥1.2) To prevent the denominator from being zero, let's solve:

{(x^{2} - 25)}^{2} \neq 0

⇥⇥

{(x^{2} - 25)}^{2} \neq 0

is equivalent to:

x^{2} - 25 \neq 0

⇥⇥The two numbers whose square is

25

are

-5

and

5

⇥⇥Solutions found:

x \notin {-5, 5}

⇥ For the expression

- \frac{3 x^{2} + 8 x + 75}{{(x^{2} - 25)}^{2}}

to be defined, we must have:

{(x^{2} - 25)}^{2} \neq 0

⇥

{(x^{2} - 25)}^{2} \neq 0

has already been solved:

x \notin {-5, 5}

⇥Domain constraints:

x \in ℝ \ {-5, 5}

⇥

f^{'} (x)

is never zero on the domain of

f

⇥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^{'} (-6) = - \frac{135}{121}

< 0:

f

is decreasing on

(-∞, -5)

⇥

f^{'} (0) = - \frac{3}{25}

< 0:

f

is decreasing on

(-5, 5)

⇥

f^{'} (6) = - \frac{21}{11}

< 0:

f

is decreasing on

(5, +∞)

⇥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

0

⇥As

x

approaches

-5

from the left,

f (x)

tends to

-∞

; from the right, to

+∞

⇥As

x

approaches

5

from the left,

f (x)

tends to

-∞

; from the right, to

+∞

⇥As

x

tends to

+∞

f (x)

tends to

0

⇥Let's draw up the variation table:

⇥

$x$	$-∞$		$-5$		$5$		$+∞$
$f^{'} (x)$		-	‖	-	‖	-
$f (x)$	$0$	↘	$-∞$ ‖ $+∞$	↘	$-∞$ ‖ $+∞$	↘	$0$

⇥So

f

is decreasing on

(-∞, -5)

and on

(-5, 5)

and on

(5, +∞)

(-∞, -5)

f

is continuous: it takes every value of

(-∞, 0)

(-5, 5)

f

is continuous: it takes every value of

ℝ

(5, +∞)

f

is continuous: it takes every value of

(0, +∞)

▶Range:

ℝ