In: Math

Explain how to solve the following problems step by step, and reason it should be solved that way:

1) Find domain for each problem and explain in brief in each case. How do you find the domain of these problems without using a graph (don't state the problem only)? Explain this: why is there a difference between in the domain of an even and an odd indexed radical? Why the domain of a function with a square root in the denominator IS different than the domain of a function with square root NOT in the denominator?

A rational function of the form

Is defined for every x except Q(x)=0. (denominator equal to zero)

So find x which corresponding to Q(x)=0. Then domain is all real number except this number.

A square root funtion

Is defined for every x except p(x) <0

So p(x) should be greater than or equal to 0.

Find x to which p(x)> 0 . This is domain.

a)

Here Q(x) is x^{2}+7 .

x^{2}+7=0

X^{2}= -7

This is not possible for any real number . So domain is entire real number .

b).

Here Q(x) is x^{2}-3x-18

x^{2}-3x -18=0

For a quadratic equation ax^{2}+bx+c=0

Here a=1 ,b= -3 ,c= -18

So

. Or.

So domain is all number except x=6 and x= -3

C)

Domain of a cube root funtion is all x .

So domain of

Is all x.

d)

This is rational function with square root funtion.

For square root funtion p(x)= 7-2x

P(x) should be greater than or equal to zero.

P(x)≥0

7-2x ≥0

Add 2x at both side . This will not effect the inequality.

7-2x+2x ≥0+2x

7≥2x

Divide both side by 2

3.5 ≥ x

So domain is all x less than 3.5 , including 3.5

Now consider condition for rational funtion.

Here

So funtion is defined for every x except

7-2x=0

2x=7

x=7/2=3.5

So funtion is not defined at x=3.5

So combining this two domain is all x less than 3.5

Domain of odd indexed radicle is all x.

Domain of even indexed radicle is only posative number.

For even

(-1)^{2}= 1^{2} =1

For odd indexed

(-1)^{3}≠ 1^{3}

They are different.

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