Answer to Question #239089 in Calculus for Rose

Question

Question 19

Identify the domain of the following functions.

a) 𝑓(𝑥) = sin−1(3𝑥 − 1)

b) 𝑓(𝑥) = [log(sin−1(√𝑥2 + 3𝑥 + 2))]

c)𝑘(𝑥)= 1 √(𝑥−2)2

d) 𝑗(𝑥)= 1 𝑥−√(𝑥+2)

e) 𝑓(𝑥)=ln|𝑥+3|−5

 

  

Answer 1

$(a) 𝑓(𝑥) = sin^{-1}(3𝑥 − 1)\\ -1<=3x-1<=1\\ \Rightarrow 0<=3x<=2\\ \Rightarrow 0<=x<=\frac{2}{3}\\ \therefore x\in [0,\frac{2}{3}]$

$(b) 𝑓(𝑥) = [log(sin^{−1}(\sqrt{𝑥^2 + 3𝑥 + 2}))]\\ \sin ^{-1} \sqrt{x^{2}+3 x+2}>0 \\ \Rightarrow \sqrt{x^{2}+3 x+2}>0 \ true \ for \ all \ \mathrm{x} . \\ Also, -1 \leq \sqrt{x^{2}+3 x+2} \leq 1\\ \Rightarrow 0<\sqrt{x^{2}+3 x+2}<=1\\ Here \ x^{2}+3 x+2>0 \quad \text { or } \ x^{2}+3 x+2 \leq 1\\ \Rightarrow x^{2}+3 x+2>0 \quad \text { or } \ x^{2}+3 x+1\leq 0\\ (x+1)(x+2)>0 \ or\ \frac{-3-\sqrt{5}}{2} \leq x \leq \frac{-3+\sqrt{5}}{2} \\ Final \ answer : {\left[\frac{-3-\sqrt{5}}{2},-2\right) \cup\left(-1, \frac{-3+\sqrt{5}}{2}\right]}$

$(c) \ 𝑘(𝑥)= \frac{1}{ \sqrt{(𝑥−2)^2}}\\ Here, \ x-2\neq0\\ \Rightarrow x\neq 2 \\ So, x\in R-{{2}}$

$(d) 𝑗(𝑥)= \frac{1}{ 𝑥−\sqrt{(𝑥+2)}}$

Here, $𝑥−\sqrt{(𝑥+2)}\neq0\\$

$\Rightarrow x\neq\sqrt{(𝑥+2)}\\ \Rightarrow x^2\neq x+2\\ \Rightarrow x^2-x-2\neq0\\ \Rightarrow (x-2)(x+1)\neq 0\\ \Rightarrow x\neq -1,2\\$

Also, 

$x+2>=0\\ x>=-2$

$\therefore x\in [-2,-1)\cup(-1,2)\cup(2,\infty)$

$(e) \ 𝑓(𝑥)=ln|𝑥+3|−5\\ Here, |𝑥+3|>0$ which is always true.

So, $x\in R$