Let f x = e x : R + → R and g x = x 2 - x : R → R , find domain and range of f ∘ g x and g ∘ f x .

f x = e x : R + → R and g x = x 2 - x : R → R f ∘ g x = f g x = e g ( x ) = e x 2 - x So, domain of f ∘ g ( x ) is R . For range, let y = x 2 - x . ⇒ y = x 2 - 2 . x . 1 2 + 1 4 - 1 4 ⇒ y = x - 1 2 2 - 1 4 So, - 1 4 ≤ x 2 - x < ∞ ⇒ e - 1 4 ≤ e x 2 - x < e ∞ Hence, range of f ∘ g ( x ) is [ e - 1 4 , ∞ ) . Now, g ∘ f x = g f x = f x 2 - f x = e 2 x - e x ; x > 0 Domain: x ∈ ( 0 , ∞ ) For x ∈ R + , g ∘ f ( x ) will be increasing function and therefore range of g ∘ f ( x ) is 0 , ∞ .

Determine f - 1 x , if a given function is invertible. f : - ∞ , - 1 → - ∞ , - 2 defined by f x = - x + 1 2 - 2 .

Let, y = f x = - x + 1 2 - 2 where f : - ∞ , - 1 → - ∞ , - 2 . ⇒ y + 2 = - x + 1 2 ⇒ x + 1 2 = - y + 2 ⇒ x + 1 = ± - y + 2 ⇒ x = - 1 ± - y + 2 Therefore, replacing y → x , ⇒ f - 1 x = - 1 ± - x + 2 ; f : - ∞ , - 2 → - ∞ , - 1 as inverse of a function domain becomes co-domain and co-domain becomes domain. Here, we see that, f - 1 x = - 1 + - x + 2 is not defined at x ∈ - ∞ , - 2 for the given co-domain. Hence, f - 1 x = - 1 - - x + 2 .

Find the domain of y = cot - 1 x 2 - 1

cot - 1 x , x ∈ R so, for y = cot - 1 x 2 - 1 x 2 - 1 ∈ R ⇒ x 2 - 1 ≥ 0 ⇒ x - 1 x + 1 ≥ 0 i.e., x ∈ − ∞ , − 1 ∪ 1 , ∞

Find the range of sin - 1 x 2 + x + 1 .

For range of sin - 1 x 2 + x + 1 , x 2 + x + 1 ≥ 0 . x + 1 2 2 + 3 4 ≥ 0 , which is true for all value of x , and minimum value is 3 4 . Therefore, 3 4 ≤ x + 1 2 2 + 3 4 < ∞ 3 4 ≤ x 2 + x + 1 < ∞ 3 2 ≤ x 2 + x + 1 < ∞ sin - 1 3 2 ≤ sin - 1 x 2 + x + 1 < sin - 1 ∞ But for sin - 1 x , value is always less than or equal to π 2 . Hence, sin - 1 3 2 ≤ sin - 1 x 2 + x + 1 ≤ π 2 π 3 ≤ sin - 1 x 2 + x + 1 ≤ π 2 i.e., sin - 1 x 2 + x + 1 ∈ π 3 , π 2 .

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Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1

Embibe Experts Solutions for Chapter: Functions, Exercise 1: Exercise-1

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Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 61

Examine whether the following pair of functions are identical or not, $f (x) = sgn (x)$ and $g (x) = \{\begin{matrix} \frac{x}{|x|} x \neq 0 \\ 0 x = 0 \end{matrix}$ .