Elementary functions – Answers

Ex.1. (domain)

(1) ⟨−1, 1⟩
(2) 0, π2 ∪ ∪k≥1 ⟨kπ, π2 + kπ )

In this case we have the following assumptions:

ˆ x > 0, because log x must be well-defined,
ˆ x≠ π2 + k · pi, where k ∈ Z, because tan x must be well-defined,
√
ˆ tan x ⩾ 0, because tan x must be well-defined,
ˆ x2 + 3x + 5 ̸= 0, because denominator can not be equal to 0.
First assumption does not require any extra calculations, so x > 0.
Second assumption also does not require any extra calculations, so x ̸=
π
2
+ k · pi, where k ∈ Z.
Third assumption requires knowledge of the function tangent; please check
the graph of it, from it
you will deduce that tan x is nonnegative if and
only if x ∈ ⟨kπ, π2 + kπ for some k ∈ Z.
To deal with the fourth assumption we check the zeros of x2 +3x+5. Since
∆ = −11 < 0 and a > 0, the expression x2 + 3x + 5 is always positive.
Then, x ∈ Df if and only if x satisfies
 π π  π 
x > 0 ∧ x ∈ ∪k∈Z − + kπ, + kπ ∧ x ∈ ∪k∈Z ⟨ kπ, + kπ ∧ x ∈ R.
2 2 2
The last condition is useless, so
 π π  π 
x > 0 ∧ x ∈ ∪k∈Z − + kπ, + kπ ∧ x ∈ ∪k∈Z ⟨ kπ, + kπ .
2 2 2
The third condition is ”contained” in the second one, so we leave only the
third one:
π 
x > 0 ∧ x ∈ ∪k∈Z ⟨ kπ, + kπ .
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Now combining the last two conditions we have that either k = 0 or k > 0.
For k = 0 we have x ∈ 0, π2 .

For k = 1 we have x ∈ ∪k≥1 ⟨kπ, π2 + kπ ).

Thus, Df = 0, π2 ∪ ∪k≥1 ⟨kπ, π2 + kπ ).


√ √
(3) (−∞, − 3 2 2 ⟩ ∪ ⟨ 3 2 2 , ∞)
(4) R \ π4 + kπ : k ∈ Z


(5) π9 + 2kπ , 5π + 2kπ

3 9 3
, k∈Z
(6) (−∞, −1⟩ ∪ (1, ∞)
(7) ⟨4 21 , ∞ )
(8) ⟨−20, 5⟩
(9) ⟨− 13 , 1⟩
(10) ⟨−4, 4⟩ \ {0}
(11) ⟨ 21 , 3⟩
(12) ⟨−1, 1⟩ \ {0}
The assumptions are the following.
ˆ arcsin(x2 ) ̸= 0
ˆ −1 ⩽ x2 ⩽ 1
From the first one we get

arcsin(x2 ) ̸= 0 ⇔ x2 ̸= 0 ⇔ x ̸= 0,

whereas from the second one

−1 ⩽ x2 ⩽ ⇔ x2 ⩽ 1 ⇔ x2 −1 ⩽ 0 ⇔ (x−1)(x+1) ⩽ 0 ⇔ x ∈ ⟨−1, 1⟩.

Summing up, Df = ⟨−1, 1⟩ \ {0}.

(13) ⟨1, 3⟩ \ {2}
(14) ⟨−13, 12⟩ \ {0}
(15) ⟨0, 4⟩ \ 13


(16) ⟨−1, 0⟩
(17) (−∞, −1⟩
(18) ⟨2, ∞)

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(19) ⟨e−6 , e−4 ⟩

(20) ⟨e, e3 ⟩
(21) (0, 3⟩
(22) (1, e2 ⟩
The assumptions:
ˆ x > 0,
ˆ −1 ⩽ 1 − ln x ⩽ 1,
ˆ arccos(1 − ln x) ̸= 0.
From the second one we get

−1 ⩽ 1 − ln x ⩽ 1 ⇔ −1 ⩽ ln x − 1 ⩽ 1 ⇔ 0 ⩽ ln x ⩽ 2
ln 1 ⩽ ln x ⩽ ln e2 ⇔ 1 ⩽ x ⩽ e2 .

From the third one we have

arccos(1 − ln x) ̸= 0 ⇔ 1 − ln x ̸= 1 ⇔ ln x ̸= 0 ⇔ ̸= x ̸= 1.

From the underlined results we get Df = (1, e2 ⟩.

√ √ √ √
(23) ⟨− 5, − 3⟩ ∪ ⟨ 3, 5⟩
√ √
6 6
(24) ⟨− 2
, 2
⟩
1
(25) (−∞, 3 2 ⟩
(26) (−∞, 3 12 ⟩
(27) ⟨1, ∞)
(28) ⟨− 13 , 23 ⟩
(29) ⟨−3, 8⟩
(30) (− 13 , ∞) \ {0}
(31) (−∞, −1⟩ \ {−7}
(32) ∪k∈Z ⟨−π + 2kπ, 2kπ⟩
The only assumption is −1 ⩽ sin3 x + 1 ⩽ 1. We have

−1 ⩽ sin3 x + 1 ⩽ 1 ⇔ −2 ⩽ sin3 x ⩽ 0 ⇔ −1 ⩽ sin3 x ⩽ 0

−1 ⩽ sin x ⩽ 0 ⇔ −π + 2kπ ⩽ x ⩽ 2kπ.

(33) ⟨0, ∞)

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√ √
10 10
(34) (−∞, − 2
⟩ ∪⟨ 2
, ∞)
3
(35) 4
, 1⟩
(36) (1, ∞)
(37) ∅

Ex.2. (domain, 2 var.)

(1) –
(2) We have the following assumptions:
i. y + 1 ⩾ 0 ⇒ y ⩾ −1,
ii. x + 2 > 0 ⇒ x > −2,
iii. ln2 (x + 2) + 1 ̸= 0 ⇒ always true.
Df = (−2, ∞) × ⟨−1, ∞).

Figure 1: Picture to Ex.2.2.

Y
4

X
−4 −2 2 4

−2

−4

(3) –
(4) –
(5) –
(6) –
(7) –

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(8) –
(9) –
(10) –
(11) –
(12) We have:
ˆ
x3 − 8y ⩾ 0 ⇔ y ⩽ 81 x3 ,
ˆ
x − 1 ⩾ 0 ⇔ x ⩾ 1,
√ √
ˆ
1 − x − 1 > 0 ⇔ x − 1 < 1 ⇔ 0 ⩽ x − 1 < 1 ⇔ 1 ⩽ x < 2,
ˆ
64 − x2 − y 2 > 0 ⇔ x2 + y 2 < 82 .
Df = (x, y) ∈ R2 : y ⩽ 18 x3 , 1 ⩽ x < 2, x2 + y 2 < 82 .


Figure 2: Picture to Ex.2.12.

10
Y

X
−10 −5 5 10

−5

−10

(13) –
(14) –
(15) –
(16) –
(17) –
(18) –

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(19) –

Ex.3. (calc.)
2π
(1) 3  
1 2π 2π
arccos − + arcsin 0 = +0=
2 3 3
(2) − 7π
12
13π
(3) 12
(4) − π6
9π 2
(5) 4
1
(6) 4
√
7
(7) 7
√
(8) 5−π
By the definition
√  √ π π


arctan(tan√ 5) = α ⇔ tan 5 = tan α ∧ α√∈ − ,
2 2
α = 5 + kπ ∧ α ∈ − π2 , π2


⇔ x = 5 − π.

(9) 4π − 12
(10) 100 − 32π
√
3
(11) 3
1
(12) 2
√
2 2
(13) 3
√
5
(14) 3
√
(15) −377
√
5
(16) 5
√
3 10
(17) 10
√
2(2+3 6)
(18) 25
1
(19) 3
√
2
(20) 2

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Ex.4. (graph)

(1) –
(2) We have the following sequence of the graphs of the functions: y = log2 x,
y = | log2 x|, y = | log2 (x − 1)| − 1, y = f (x), which are depicted below:

Figure 3: Graphs to example 2 from Ex.4.

10
Y y = log2 x
y = | log2 x|
8
y = | log2 (x − 1)| − 1
y = || log2 (x − 1)| − 1|
6

−5 5 10 15 20

−2

−4

−6

−8

−10

(3) –
(4) –
(5) –
(6) –
(7) –
(8) –
(9) –
(10) –
(11) –
(12) –

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Figure 4: Graphs to example 2 from Ex.4.

3
Y y = arccos x
y = arccos x2
y = arccos | x
2
|

2 y = arccos | x+1
2
|− π
4

y = arccos | x+1
2
|− π
4

−3 −2 −1 1 2 3

−1

−2

−3

(13) We have the following sequence of the graphs of the functions: y =

arccos x, y = arccos x2 , y = arccos x2 , y = arccos x+1
2
− π4 , y = arccos x+1
2
− π
4
.

Ex.5. (parity)

(1) even
Df = ⟨−1, 1⟩ – such domain can be a domain of an even or odd function,
so we check f (−x):
arcsin
f (−x) = (−x) · arcsin(−x) odd
= −x · (− arcsin x) = x arcsin x = f (x).

Since f (−x) = f (x) for all x ∈ Df , f is even.

(2) neither even nor odd
(3) neither even nor odd
(4) even

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(5) odd
(6) even
(7) even
(8) odd
(9) neither even nor odd

Ex.6. (proof, inj.)

(1) –
(2) Let x1 , x2 ∈ Df = R. Suppose that f (x1 ) = f (x2 ). We have then
3 3
2x1 −1 = 2x2 −1 ⇔ x31 − 1 = x32 − 1 ⇔ x31 = x32 ⇔ x1 = x 2 ,

so f is injective.
(3) –

Ex.7. (proof, incr.)

(1) –
(2) Suppose that x1 , x2 ∈ Df = (0, ∞) and x1 < x2 . Then
2· inscreasing
3x1 < 3x2 ⇒ 23x1 < 23x2 ⇒ 23x1 − 1 < 23x2 − 1
ln(·) increasing
⇒ ln(23x1 − 1) < ln(23x2 − 1) ⇒ f (x1 ) < f (x2 ),

so f is increasing.
(3) –

Ex.8. (proof, decr.)

(1) –
(2) Let x1 , x2 ∈ Df = R and x1 < x2 . Then

·3 , ·7 increasing
⇒ x31 < x32 , x71 < x72 ⇒ x31 < x32 , 2x71 < 2x72 ⇒ 2x71 + x31 < 2x72 + x32
e· increasing 7 3 7 3 7 3 7 3
⇒ e2x1 +x1 < e2x2 +x2 ⇒ −e2x2 +x2 < −e2x1 +x1 ⇒ f (x2 ) < f (x1 ),

so f is decreasing.
(3) –

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Ex.9. (inverse funct.)

√
(1) Df = R, f −1 (x) = 3
x − 1, Df −1 = R
(2) Df = R \ {−2}, f −1 (x) = −2x−1
x−1
, Df −1 = \ {1}
We have x ∈ Df iff x + 2 ̸= 0, so Df = R \ {−2}.
To find the formula for the inverse function we determine x from the
x−1
relation y = x+2 :
x−1 y−1̸=0 −2y − 1
y= ⇔ yx+2y = x−1 ⇔ x(y−1) = −2y−1 ⇔ x= .
x+2 y−1

Thus, f −1 (y) = −2y−1

y−1
, i.e. f −1 (x) = −2x−1
x−1
. The domain of f −1 fol-
lows from the assumptions for y that appeared during transformations, so
Df −1 = R \ {1}.
(3) Df = R, f −1 (x) = log25x+3 , Df −1 = R+
√ √
(4) Df = (− 5 2, ∞), f −1 (x) = 5 4x − 2, Df −1 = R
(5) Df = R, f −1 does not exists
3
(6) Df = R \ {−4}, f −1 (x) = − x4x 3 −1 , Df −1 = R \ {1}
q
3 y5
(7) f −1 (x) = arctan e 2 , Df −1 = R
√
3
(8) Df = R, f −1 (x) = ln 5 tan y − 1 , Df −1 = 0, π2



−x
(9) Df = (0, ∞) \ 1e , f −1 (x) = e x−1 , Df −1 = R \ {1}


(10) Df = 1e , e , f −1 (x) = esin x , Df −1 = ⟨0, π2 ⟩

(11) Df = R \ {0}, f −1 (x) = cot x, Df −1 = − π2 , π2 \ {0}

(12) Df = (0, ∞), f −1 does not exists

Since cot is not injective, we may doubt whether f is injective. In fact,
we have for instance
2π+2 2π+2
2π+2 π−1 +2
f (3 π−1 ) = cot log3 3 2π+2 +2 = cot π−1
2π+2
−2
= cot π = 0,
log3 3 π−1 −2 π−1
log3 3−2 +2
f (3−2 ) = cot log −2 −2 = cot −2+2
−2−2
= cot 0 = 0,
33

2π+2
although 3 π−1 ̸= 3−2 , so f is not injective, and therefore, does not have
an inverse function.
(13) Df = ⟨−π, π⟩ \ {0}, f −1 (x) = sin πx , Df −1 = −∞, − 21 ⟩ ∪ ⟨ 12 , ∞

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