range of f(x)=-e^{-x} — step-by-step solution

⚠ f(x)=-e^{-x} is read as f(x)=-e^(-x): use ^(...) to group an exponent, use * to multiply

1) Let's find the real values taken by

- e^{- x}

using its variation table

⇥Let's study the variations of

f (x)

: first find where its derivative

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

is zero

⇥Let's solve in

ℝ

\frac{1}{e^{x}} = 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 avoid a zero denominator, let's solve:

e^{x} \neq 0

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

e^{x} > 0

, so

e^{x} \neq 0

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) = 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

-∞

⇥As

x

tends to

+∞

f (x)

tends to

0

⇥Let's draw up the variation table:

⇥

⇥So

f

is increasing on

ℝ

ℝ

f

is continuous: it takes every value of

(-∞, 0)

▶Range:

(-∞, 0)