f(x)=ln(x) — step-by-step solution

We save the definition:

f (x) = \ln (x)

Let's study the function

f (x) = \ln (x)

1) Let's find the domain of

\ln (x)

For the expression

\ln (x)

to exist, you must check:

x > 0

Solution found:

x > 0

Domain of definition:

x > 0

2) Let's study the parity of

\ln (x)

with respect to

x

The domain is

x > 0

This domain is not symmetric about 0: the function is neither even nor odd

Parity: neither even nor odd

3) Differentiability

Derivative:

\frac{1}{x}

4) Let's study the variations of

f (x)

: first find where its derivative

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

is zero

Let's solve in

ℝ

\frac{1}{x} = 0

⇥4.1) Let's solve in

ℝ

1 = 0

⇥The statement

1 = 0

is false, whatever the value of

x

⇥Solution:

x \in \emptyset

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

x \neq 0

For the expression

\frac{1}{x}

to exist, you must check:

x \neq 0

Domain constraints:

x \neq 0

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^{'} (1) = 1

> 0:

f

is increasing on

(0, +∞)

Let's calculate the values of

f

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

x

approaches

0

from the right,

f (x)

tends to

-∞

x

tends to

+∞

f (x)

tends to

+∞

Let's draw up the variation table:

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

f

is increasing on

(0, +∞)

5) Let's read the range from the variation table

⇥On

(0, +∞)

f

is continuous: it takes every value of

ℝ

Range:

ℝ

6) On the horizontal axis, y = 0: solve

\ln (x) = 0

Let's solve in

ℝ

\ln (x) = 0

On the logarithm's domain, ln is strictly increasing:

\ln (x) = 0

is equivalent to

x = 1

Candidates, pending the domain check:

x = 1

For the expression

\ln (x)

to be defined, we must have:

x > 0

Domain constraints:

x > 0

The graph crosses the horizontal axis at

(1, 0)

x = 0 is not in the domain: the graph does not cross the vertical axis

Axis intercepts:

(1, 0)

7) Let's find the asymptotes of

\ln (x)

, for real

x

The function is defined on

(0, +∞)

: a vertical asymptote can only be at a finite bound of this domain

x

approaches

0

from the right, the function tends to

-∞

The function tends to infinity: the line

x = 0

is a vertical asymptote

x

tends to

+∞

, the function tends to

+∞

and

\frac{f (x)}{x}

tends to

0

: no horizontal or slant asymptote at

+∞

🖼/draw?f1=ln%28x%29&v=x&x2=0&b=-10.0&e=10.0&min=-0.21972245773362198&max=2.4169470350698417&dash=2

Asymptotes:

x = 0

▶Study of $f (x)$ : Domain of definition: $x > 0$ ; Parity: neither even nor odd; Derivative: $\frac{1}{x}$ ; Variations: increasing on $(0, +∞)$ ; Range: $ℝ$ ; Axis intercepts: $(1, 0)$ ; Asymptotes: $x = 0$