range of arctan(x) — step-by-step solution

1) Let's find the real values taken by

\arctan (x)

using its variation table

⇥Let's study the variations of

f (x)

: first find where its derivative

f^{'} (x) = \frac{1}{x^{2} + 1}

is zero

⇥Let's solve in

ℝ

\frac{1}{x^{2} + 1} = 0

⇥⇥1.1) Let's solve in

ℝ

1 = 0

⇥⇥The statement

1 = 0

is false, whatever the value of

x

⇥⇥Solution:

x \in \emptyset

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

x^{2} + 1 \neq 0

⇥⇥

x^{2} \neq -1

⇥⇥In ℝ, a square is never negative: no real number has

-1

as its square

⇥⇥Solution:

x \in \emptyset

⇥

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^{'} (0) = 1

> 0:

f

is increasing on

ℝ

⇥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

- \frac{π}{2}

⇥As

x

tends to

+∞

f (x)

tends to

\frac{π}{2}

⇥Let's draw up the variation table:

⇥

⇥So

f

is increasing on

ℝ

ℝ

f

is continuous: it takes every value of

(- \frac{π}{2}, \frac{π}{2})

▶Range:

(- \frac{π}{2}, \frac{π}{2})