extreme f(x)=((x^2-1))/(x^2+1) — step-by-step solution

1) Find the strict local extrema of

\frac{x^{2} - 1}{x^{2} + 1}

: they are the points of the domain where its derivative changes sign

⇥Let's study the variations of

f (x)

: first find where its derivative

f^{'} (x) = \frac{4 x}{{(x^{2} + 1)}^{2}}

is zero

⇥Let's solve in

ℝ

\frac{4 x}{{(x^{2} + 1)}^{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

4 x = 0

⇥⇥We simplify the expression by

4

⇥⇥⇥Let's solve in

ℝ

x = 0

⇥⇥⇥Solution found:

x = 0

⇥⇥1.2) To avoid a zero denominator, let's solve:

{(x^{2} + 1)}^{2} \neq 0

⇥⇥

{(x^{2} + 1)}^{2} \neq 0

means that:

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 zero at

x = 0

⇥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 decreasing on

(-∞, 0)

⇥

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

⇥As

x

tends to

-∞

f (x)

tends to

1

⇥

f (0) = \frac{0^{2} - 1}{0^{2} + 1} = -1

⇥As

x

tends to

+∞

f (x)

tends to

1

⇥Let's draw up the variation table:

⇥

⇥So

f

is decreasing on

(-∞, 0]

and increasing on

[0, +∞)

x = 0

f

' changes from - to +:

f

has a strict local minimum,

f (0) = -1

▶Strict local extrema:

{(0, -1)}