Step-by-step explanation:
- Domain is the set of all possible input values for which the function is defined.
- Range is the set of all possible output values after we have substitute all the possible values in the domain.
Question # 1 Solution
Determining the domain and range of [tex]f(x, y)=(x,y, x-y)[/tex]
Given the function
[tex]f(x, y)=(x,y, x-y)[/tex]
As it is clear that for whatever the value of x we put in the given function, the function will be defined for all real numbers. Hence, the domain will be the set of all real numbers i.e. [tex]\:\left(-\infty \:,\:\infty \:\right)[/tex].
Also, we it is clear that for whatever the value we input, we get the output value ranging from the set of all the real numbers. So, the range will be the set of all real numbers.
Therefore,
Domain: [tex]\:\left(-\infty \:,\:\infty \:\right)[/tex]
Range: [tex]\:\left(-\infty \:,\:\infty \:\right)[/tex]
Question # 2 Solution
Determining the domain and range of [tex]f(x) = x^{2} -3[/tex]
As it is clear that for whatever the value of x we put in the given function i.e.[tex]f(x) = x^{2} -3[/tex], the function will be defined for all real numbers. Hence, the domain of [tex]f(x) = x^{2} -3[/tex] will be the set of all real numbers i.e. [tex]\:\left(-\infty \:,\:\infty \:\right)[/tex].
Also, we it is clear that for whatever the value we input, we get the output value ranging from -3 to positive infinity. So, the range will be [tex]\:[-3,\:\infty \:)[/tex] .
Therefore,
Domain: [tex]\:\left(-\infty \:,\:\infty \:\right)[/tex]
Range: [tex]\:[-3,\:\infty \:)[/tex]
The graph is also attached in figure a.
Question # 3 Solution
Determining the domain and range of [tex]f(x) =\frac{1}{\sqrt{1-x^{2} } }[/tex]
As the domain of a function is the set of input or argument values for which the function is real and defined.
As we know that the value inside square root can not be defined.
i.e.
[tex]\sqrt{f\left(x\right)}\quad \Rightarrow \quad \:f\left(x\right)\ge 0\:[/tex]
So,
[tex]1-x^2\ge \:0[/tex]
[tex]-x^2\ge \:-1[/tex]
[tex]x^2\le \:1[/tex]
[tex]\mathrm{For\:}u^n\:\le \:\:a\mathrm{,\:if\:}n\:\mathrm{is\:even}\mathrm{\:then\:}-\sqrt[n]{a}\:\le \:\:u\:\le \sqrt[n]{a}[/tex]
[tex]-1\le \:x\le \:1[/tex]
So, domain will be: [tex]\:\left(-1,\:1\right)[/tex]
Finding the range for the interval [tex]\:\left(-1,\:1\right)[/tex].
[tex]\mathrm{The\:interval\:has\:a\:minimum\:point\:at\:}x=0\:\mathrm{\:with\:value\:}f\left(0\right)=1[/tex]
[tex]\mathrm{Combine\:the\:function\:value\:at\:the\:edge\:with\:the\:extreme\:points\:of\:the\:function\:in\:the\:interval:}[/tex][tex]\mathrm{Minimum\:function\:value\:at\:the\:domain\:interval\:}-1<x<1\mathrm{\:is\:}1[/tex]
[tex]\mathrm{Maximum\:function\:value\:at\:the\:domain\:interval\:}-1<x<1\mathrm{\:is\:}\infty \:[/tex]
[tex]\mathrm{Therefore\:the\:range\:of\:}\frac{1}{\sqrt{1-x^2}}\:\mathrm{\:at\:the\:domain\:interval\:}-1<x<1\:\mathrm{\:is}[/tex]
[tex]1\le \:f\left(x\right)<\infty \:[/tex]
[tex]\mathrm{Combine\:the\:ranges\:of\:all\:domain\:intervals\:to\:obtain\:the\:function\:range}[/tex]
[tex]f\left(x\right)\ge \:1[/tex]
So,
Range: [tex][1,\:\infty \:)[/tex]
The graph is also attached in figure b.
Keywords: domain, range
Learn more about domain and range from brainly.com/question/12918386
#learnwithBrainly
