Find Domain of Each Function Part 2

Section 1.4
Questions
52, 54, 56

See attachment.

52. Set denominator = 0 and solve for x

x^2 - 4x

x(x - 4) = 0

x = 0

x - 4 = 0

x = 4

Domain = All real numbers except x cannot be 0 and 4.

54. Set denominator to 0 and solve for x.

6 + x = 0

x = -6

Set radicand to be => 0.

x + 6 => 0

x => -6

Domain = All real numbers except that x > -6.

56. Set denominator to be => 0.

x - 10 => 0

x => 10

Set numerator = 0 and solve for x.

x + 2 = 0

x = - 2

Domain = All real numbers except x cannot be -2 and 10.

 

52 and 54 are correct

56.
domain
{x element R : x>10}

 

Can you explain why my domain answer for 56 is wrong?

 

56.

f(x)=(x+2)/sqrt(x-10)
you say: Domain = All real numbers except x cannot be -2 and 10.
-2 could be used because numerator can be equal to zero
denominator is sqrt(x-10) and cannot be equal to zero; so, x=10 is excluded from domain
also, all real numbers less than zero are excluded because any of them will make sqrt of negative number which does not have real solution

therefore, domain is
{x element R : x>10}

 

I understood what you said but some extra questions for you.

1. What does {x element R : x>10}?

2. It's ok for x to be -2 because zero in numerator is acceptable. If this is the case, then why x > 10? In x > 10, -2 is excluded.

3. Domain = ALL REAL NUMBERS except that x must be greater than 10. You say?

 

1. What does {x element R : x>10}?
it says domain is all real numbers x such that x>10,
or domain is all real numbers x greater than 10

2. -2 is excluded because of denominator is sqrt(x-10) => if you use -2 you get sqrt(-2-10)=sqrt(-12) which is imaginary solution and we are looking for a domain that will give as real solution

3. Domain = ALL REAL NUMBERS except that x must be greater than 10.
not except that,
Domain = ALL REAL NUMBERS where x is greater than 10.

 

You said:

"-2 is excluded because of denominator is sqrt(x-10) => if you use -2 you get sqrt(-2-10)=sqrt(-12) which is imaginary solution and we are looking for a domain that will give as real solution."

Is there a math class that accepts imaginary solutions? If so, what is the name of the course?

 

in this case -2 is solution for function above, but imaginary one ( only the function is not real)

since you need to define domain, recall definition of it:

The domain of a function is the set of input or argument values for which the function is real and defined.

for better understanding, see the graph

 

I will post a few questions like 56 to get additional practice. It's important for me to get a solid understanding of domain (and later on the range) of functions. Otherwise, the rest of precalculus will be a nightmare, to say the least.