Domain of A Function

[nycmathguy]

Domain:

The domain of a function is the set of all possible inputs for the function.

Find the domain for each function below.

48. g(x) = 1 - x^2

For 48, I say the domain is all real numbers.

50. f(t) = (t + 4)^(1/3)

For 50, I must set the radicand to be greater than or equal to 0.

t + 4 => 0

t => - 4

I will say the domain is can be all real numbers except for -4. Yes?

52. h(x) = 6/(x^2 - 4x)

Set denominator to 0 and solve for x.

x^2 - 4x = 0

x(x - 4) = 0

x = 0

x - 4 = 0

x = 4

Domain: All real numbers except for x = 0
and x = 4.

You say?

54. f(x) = sqrt{x + 6}/(6 + x)

I am not too sure about this one considering that the function has a square root numerator.

56. f(x) = (x + 2)/sqrt{ x - 10}

I will set denominator to be greater than or equal to 0.

x - 10 => 0

x => 10

This does not make sense. Clearly, x must be greater than 10. Here, x cannot be 10. Division by 0 is undefined.

Domain: x > 10

You say?

 

[MathLover1]

Domain:

The domain of a function is the set of all possible inputs for the function.

Find the domain for each function below.

48. g(x) = 1 - x^2

For 48, I say the domain is all real numbers.

50. f(t) = (t + 4)^(1/3)

For 50, I must set the radicand to be greater than or equal to 0.

t + 4 => 0

t => - 4

I will say the domain is can be all real numbers except for -4. Yes?

52. h(x) = 6/(x^2 - 4x)

Set denominator to 0 and solve for x.

x^2 - 4x = 0

x(x - 4) = 0

x = 0

x - 4 = 0

x = 4

Domain: All real numbers except for x = 0
and x = 4.

You say?

54. f(x) = sqrt{x + 6}/(6 + x)

I am not too sure about this one considering that the function has a square root numerator.

56. f(x) = (x + 2)/sqrt{ x - 10}

I will set denominator to be greater than or equal to 0.

x - 10 => 0

x => 10

This does not make sense. Clearly, x must be greater than 10. Here, x cannot be 10. Division by 0 is undefined.

Domain: x > 10

You say?

For 48 your answer is correct


For 50 you got t >= - 4, so domain is

all real numbers t element R where t>=-4 (recall f(t) = (t + 4)^(1/3) is third root of t+4, and if t=-4 you get third root of 0 which is zero)

for 52 your answer is correct

for 54. f(x) = sqrt(x + 6)/(6 + x)
(6 + x)=0 if x=-6

sqrt(x + 6) =0 if x=-6
so domain is all real numbers x element R : x>-6

56. your answer is correct

 

[nycmathguy]

For 48 your answer is correct


For 50 you got t >= - 4, so domain is

all real numbers t element R where t>=-4 (recall f(t) = (t + 4)^(1/3) is third root of t+4, and if t=-4 you get third root of 0 which is zero)

for 52 your answer is correct

for 54. f(x) = sqrt(x + 6)/(6 + x)
(6 + x)=0 if x=-6

sqrt(x + 6) =0 if x=-6
so domain is all real numbers x element R : x>-6

56. your answer is correct

Do you want to stay here or the other math forum?

 

[MathLover1]

Do you want to stay here or the other math forum?

I like it so far, and I want to learn these codes to type equations, fractions, and other things

 

[nycmathguy]

I like it so far, and I want to learn these codes to type equations, fractions, and other things

In case you didn't get the other reply, here is a smaller version.

A. Let's stay here for the rest of precalculus.

B. Use the other site for college algebra, which coincides with precalculus.

C. Studying both courses will greatly polish my algebra skills needed for calculus.

You say?

 

[nycmathguy]

For 48 your answer is correct


For 50 you got t >= - 4, so domain is

all real numbers t element R where t>=-4 (recall f(t) = (t + 4)^(1/3) is third root of t+4, and if t=-4 you get third root of 0 which is zero)

for 52 your answer is correct

for 54. f(x) = sqrt(x + 6)/(6 + x)
(6 + x)=0 if x=-6

sqrt(x + 6) =0 if x=-6
so domain is all real numbers x element R : x>-6

56. your answer is correct

Can you explain 50 in a different way?

 

[nycmathguy]

For 48 your answer is correct


For 50 you got t >= - 4, so domain is

all real numbers t element R where t>=-4 (recall f(t) = (t + 4)^(1/3) is third root of t+4, and if t=-4 you get third root of 0 which is zero)

for 52 your answer is correct

for 54. f(x) = sqrt(x + 6)/(6 + x)
(6 + x)=0 if x=-6

sqrt(x + 6) =0 if x=-6
so domain is all real numbers x element R : x>-6

56. your answer is correct

You said this concerning 54:

"so domain is all real numbers x element R : x>-6"

What does "x element R : x>-6" mean? Please, reply in layman terms. If you are going to use textbook math jargon, it will defeat the purpose for taking the precalculus journey with mme. Understand? I hope you and jonah have no issues.

 

[MathLover1]

You said this concerning 54:

"so domain is all real numbers x element R : x>-6"

What does "x element R : x>-6" mean? Please, reply in layman terms. If you are going to use textbook math jargon, it will defeat the purpose for taking the precalculus journey with mme. Understand? I hope you and jonah have no issues.

here you have f(x) = sqrt(x + 6)/(6 + x) and domain of the function is all real numbers except those that make denominator equal to zero

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

if x+6=0 =>will be only if x=-6
so, -6 is excluded from domain

and domain is all real numbers x element R except x=-6

but, since we have sqrt(x + 6) in numerator we know that the function is real and defined only for x values that are x>=-6
so x>=-6

since denominator excluded -6 from domain, x is not >=-6 anymore, it's just x>-6

entire function f(x) = sqrt(x + 6)/(6 + x) has a domain all real numbers x element R :
x>-6
that means domain is x=-5,-4,-3,........1,2,3,...........,infinity

in interval notation: (-6,infinity)

 

[nycmathguy]

here you have f(x) = sqrt(x + 6)/(6 + x) and domain of the function is all real numbers except those that make denominator equal to zero

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

if x+6=0 =>will be only if x=-6
so, -6 is excluded from domain

and domain is all real numbers x element R except x=-6

but, since we have sqrt(x + 6) in numerator we know that the function is real and defined only for x values that are x>=-6
so x>=-6

since denominator excluded -6 from domain, x is not >=-6 anymore, it's just x>-6

entire function f(x) = sqrt(x + 6)/(6 + x) has a domain all real numbers x element R :
x>-6
that means domain is x=-5,-4,-3,........1,2,3,...........,infinity

in interval notation: (-6,infinity)

Thank you very much. I have a question for you.
Do you think it is important to read the chapter notes leading to the questions at the end of each section? I was heavily criticized by jonah and romsek at the FORMER site for skipping the chapter notes. The problem is that, as you know well, most math books are not written in a student-friendly manner. The English is easy but the lesson itself is not so easy to understand and less easy to explain.

You say?

 

[MathLover1]

Thank you very much. I have a question for you.
Do you think it is important to read the chapter notes leading to the questions at the end of each section? I was heavily criticized by jonah and romsek at the FORMER site for skipping the chapter notes. The problem is that, as you know well, most math books are not written in a student-friendly manner. The English is easy but the lesson itself is not so easy to understand and less easy to explain.

You say?

it is crucial to read and (get good understanding) the chapter AND notes leading to the questions
without that is impossible to solve the problems

every student can make a book "student-friendly manner"
when you read one lesson, read slowly and underline key words, definitions, formulas
then take notes
then start with problem solving
read problem carefully, pay attention to what is given and what do you need to find
then go back to notes, and choose what is needed to solve that problem