Chapter 1.4, Problem 2QR

Solution

(1)

To find: Domain of $f(x)=\frac{x-2}{x+3}$

Domain of $f(x)$ is $(-\infty ,-3)\cup (-3,\infty )$

Concept used: Domain of a function is the set of all inputs on which function is defined successfully.

Calculation:

The function $f(x)=\frac{x-2}{x+3}$ is not defined when the denominator is zero.

That is, when $x+3=0$

  $\Rightarrow x=-3$

That is, $f(x)$ is not defined at $x=-3$ .

Therefore, Domain of $f(x)$ is $(-\infty ,-3)\cup (-3,\infty )$

(2)

To find: Domain of $g(x)=\ln (x-1)$

Domain of $g(x)$ is $\left(1,\infty \right)$

Concept used: Domain of a function is the set of all inputs on which function is defined successfully.

Calculation:

The function $g(x)=\ln (x-1)$ is not defined when $x-1=0$ .

That is, at $x=1$ given function is not defined.

Therefore, Domain of $g(x)$ is $\left(1,\infty \right)$

(3)

To find: Domain of $f(t)=\sqrt{5-t}$

Domain of $f(t)$ is $\left(-\infty ,5\right)$

Concept used: Function $f(x)=\sqrt{x}$ is defined only when $x>0$

Calculation:

The function $f(t)=\sqrt{5-t}$ is defined only when $5-t>0\Rightarrow 5>t$

That is, $f(t)$ is defined when $t<5$

Therefore, Domain of $f(t)$ is $\left(-\infty ,5\right)$

(4)

To find: Domain of $g(x)=\frac{3}{\sqrt{2x-1}}$

Domain of $g(x)$ is $\left(0.5,\infty \right)$

Concept used: Function $f(x)=\sqrt{x}$ is defined only when $x>0$

Calculation:

The function $g(x)=\frac{3}{\sqrt{2x-1}}$ is defined only when $2x-1>0\Rightarrow 2x>1$

That is, $g(x)$ is defined when $x>\frac{1}{2}$

Therefore, Domain of $g(x)$ is $\left(0.5,\infty \right)$

(5)

To find: Domain of $f(x)=\sqrt{\ln x}$

Domain of $f(x)$ is $\left[1,\infty \right)$

Concept used: Function $f(x)=\sqrt{x}$ is defined only when $x>0$

Calculation:

The function $f(x)=\sqrt{\ln x}$ is defined only when $\ln x>0\Rightarrow x\ge 1$

That is, $f(x)$ is defined when $x\ge 1$

Therefore, Domain of $f(x)$ is $\left[1,\infty \right)$

(6)

To find: Domain of $h(x)=\sqrt{1-x^2}$

Domain of $h(x)$ is $\left[-1,1\right]$

Concept used: Function $f(x)=\sqrt{x}$ is defined only when $x>0$

Calculation:

The function $h(x)=\sqrt{1-x^2}$ is defined only when $1-x^2>0\Rightarrow 1>x^2$

That is, $h(x)$ is defined when $-1<x<1$

Therefore, Domain of $h(x)$ is $\left[-1,1\right]$

(7)

To find: Domain of $f(t)=\frac{t+5}{t^2+1}$

Domain of $f\left(t\right)$ is $\left[-1,1\right]$

Calculation:

The function $f(t)=\frac{t+5}{t^2+1}$ is defined only when $t^2+1\ne 0\Rightarrow t^2\ne -1$

But $t^2$ is always a positive number , for all t

Therefore, Domain of $f\left(t\right)$ is $\left[-\infty ,\infty \right]$

(8)

To find: Domain of $g(t)=\ln (\left|t\right|)$

Domain of $g\left(t\right)$ is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$

Calculation:

The function $g(t)=\ln (\left|t\right|)$ is defined only when $\left|t\right|>0$

Therefore, Domain of $g\left(t\right)$ is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$

(9)

To find: Domain of $f(x)=\frac{1}{\sqrt{1-x^2}}$

Domain of $f\left(x\right)$ is $\left(-1,1\right)$

Calculation:

The function $f(x)=\frac{1}{\sqrt{1-x^2}}$ is defined only when $1-x^2>0\Rightarrow 1>x^2$

Therefore, Domain of $f\left(x\right)$ is $\left(-1,1\right)$

(10)

To find: Domain of $g(x)=2$

Domain of $g\left(x\right)$ is $\left(-\infty ,\infty \right)$

Calculation:

The function $g(x)=2$ is independent of $x$ , and is defined for all $x$

Therefore, Domain of $g\left(x\right)$ is $\left(-\infty ,\infty \right)$