extreme f(x)=(e^x)/(5+e^x) — step-by-step solution

1) Find the strict local extrema of

\frac{e^{x}}{5 + e^{x}}

: 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{5 e^{x}}{{(e^{x} + 5)}^{2}}

is zero

⇥Let's solve in

ℝ

\frac{5 e^{x}}{{(e^{x} + 5)}^{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

5 e^{x} = 0

⇥⇥We simplify the expression by

5

⇥⇥⇥Let's solve in

ℝ

e^{x} = 0

⇥⇥⇥In ℝ, the exponential function is strictly positive:

e^{x} > 0

, so

e^{x} = 0

has no solution

⇥⇥⇥Solution:

x \in \emptyset

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

{(e^{x} + 5)}^{2} \neq 0

⇥⇥

{(e^{x} + 5)}^{2} \neq 0

means that:

e^{x} + 5 \neq 0

⇥⇥

e^{x} \neq -5

⇥⇥In ℝ, the exponential function is strictly positive:

e^{x} > 0

, so

e^{x} \neq -5

is true for every

x

⇥⇥Solution obtained:

x \in ℝ

⇥

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) = \frac{5}{36}

> 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

0

⇥As

x

tends to

+∞

f (x)

tends to

1

⇥Let's draw up the variation table:

⇥

⇥So

f

is increasing on

ℝ

The derivative changes sign at no point of the domain: there is no strict local extremum

▶Strict local extrema:

\emptyset