QuestionDetermine the domain and range of the function $f(u)=\frac{u+1}{1+\frac{1}{u+1}}$ .

Studdy Solution

STEP 1

Assumptions1. The function is $f(u)=\frac{u+1}{1+\frac{1}{u+1}}$ . We need to find the domain and range of this function

STEP 2

The domain of a function is the set of all possible input values (in this case, 'u') which will make the function "work", and will output real numbers.First, let's find the domain of the function. We need to find for which values of 'u' the function is defined.The function will be undefined when the denominator of the fraction is zero. So, we need to find the values of 'u' for which the denominator is not zero.
The denominator of the function is $1+\frac{1}{u+1}$ .

STEP 3

Set the denominator equal to zero and solve for 'u'.
$1+\frac{1}{u+1} =0$

STEP 4

Rearrange the equation to isolate the term with 'u'.
$\frac{1}{u+1} = -1$

STEP 5

olve for 'u' by multiplying both sides by $u+1$ .
$u+1 = -1$

STEP 6

Subtract1 from both sides to solve for 'u'.
$u = -1 -1$

STEP 7

Calculate the value of 'u'.
$u = -2$ So, the function is undefined for $u = -2$ . Therefore, the domain of the function is all real numbers except $-2$ .

STEP 8

Now, let's find the range of the function. The range of a function is the set of all possible output values (in this case, 'f(u)'), which we get after substituting all the possible values of 'u' from the domain into the function.
To find the range, we can simplify the function and then find the values that 'f(u)' can take.

STEP 9

implify the function by multiplying the numerator and the denominator by $u+$ .
$f(u) = \frac{(u+)^2}{(u+) +}$

STEP 10

implify the function further.
$f(u) = \frac{u^2 +2u +}{u +2}$

STEP 11

Now, let's find the values that 'f(u)' can take. We can see that 'f(u)' can take all real values except for when the denominator is zero.Set the denominator equal to zero and solve for 'f(u)'.
$u + =0$

STEP 12

Subtract2 from both sides to solve for 'u'.
$u = -2$ So, 'f(u)' is undefined for $u = -2$ . Therefore, the range of the function is all real numbers except the value of the function at $u = -2$ .

STEP 13

Find the value of the function at $u = -2$ .
$f(-2) = \frac{(-2)^2 +2(-2) +}{-2 +2}$

STEP 14

implify the function to find the value.
$f(-2) = \frac{4 -4 +}{0}$ The value of the function at $u = -2$ is undefined. Therefore, the range of the function is all real numbers.
In conclusion, the domain of the function $f(u)=\frac{u+}{+\frac{}{u+}}$ is all real numbers except $-2$ , and the range of the function is all real numbers.

Was this helpful?

QuestionDetermine the domain and range of the function $f(u)=\frac{u+1}{1+\frac{1}{u+1}}$ .

Studdy Solution

STEP 1

Assumptions1. The function is $f(u)=\frac{u+1}{1+\frac{1}{u+1}}$ . We need to find the domain and range of this function

STEP 2

STEP 3

Set the denominator equal to zero and solve for 'u'.
$1+\frac{1}{u+1} =0$

STEP 4

Rearrange the equation to isolate the term with 'u'.
$\frac{1}{u+1} = -1$

STEP 5

olve for 'u' by multiplying both sides by $u+1$ .
$u+1 = -1$

STEP 6

Subtract1 from both sides to solve for 'u'.
$u = -1 -1$

STEP 7

Calculate the value of 'u'.
$u = -2$ So, the function is undefined for $u = -2$ . Therefore, the domain of the function is all real numbers except $-2$ .

STEP 8

STEP 9

implify the function by multiplying the numerator and the denominator by $u+$ .
$f(u) = \frac{(u+)^2}{(u+) +}$

STEP 10

implify the function further.
$f(u) = \frac{u^2 +2u +}{u +2}$

STEP 11

Now, let's find the values that 'f(u)' can take. We can see that 'f(u)' can take all real values except for when the denominator is zero.Set the denominator equal to zero and solve for 'f(u)'.
$u + =0$

STEP 12

Subtract2 from both sides to solve for 'u'.
$u = -2$ So, 'f(u)' is undefined for $u = -2$ . Therefore, the range of the function is all real numbers except the value of the function at $u = -2$ .

STEP 13

Find the value of the function at $u = -2$ .
$f(-2) = \frac{(-2)^2 +2(-2) +}{-2 +2}$

STEP 14

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionDetermine the domain and range of the function f(u)=u+11+1u+1f(u)=\frac{u+1}{1+\frac{1}{u+1}}f(u)=1+u+11​u+1​.

Studdy Solution

STEP 1

Assumptions1. The function is f(u)=u+11+1u+1f(u)=\frac{u+1}{1+\frac{1}{u+1}}f(u)=1+u+11​u+1​ . We need to find the domain and range of this function

STEP 2

STEP 3

Set the denominator equal to zero and solve for 'u'.1+1u+1=01+\frac{1}{u+1} =01+u+11​=0

STEP 4

Rearrange the equation to isolate the term with 'u'.1u+1=−1\frac{1}{u+1} = -1u+11​=−1

STEP 5

olve for 'u' by multiplying both sides by u+1u+1u+1.u+1=−1u+1 = -1u+1=−1

STEP 6

Subtract1 from both sides to solve for 'u'.u=−1−1u = -1 -1u=−1−1

STEP 7

Calculate the value of 'u'.u=−2u = -2u=−2So, the function is undefined for u=−2u = -2u=−2. Therefore, the domain of the function is all real numbers except −2-2−2.

STEP 8

STEP 9

implify the function by multiplying the numerator and the denominator by u+u+u+.f(u)=(u+)2(u+)+f(u) = \frac{(u+)^2}{(u+) +}f(u)=(u+)+(u+)2​

STEP 10

implify the function further.f(u)=u2+2u+u+2f(u) = \frac{u^2 +2u +}{u +2}f(u)=u+2u2+2u+​

STEP 11

Now, let's find the values that 'f(u)' can take. We can see that 'f(u)' can take all real values except for when the denominator is zero.Set the denominator equal to zero and solve for 'f(u)'.u+=0u + =0u+=0

STEP 12

Subtract2 from both sides to solve for 'u'.u=−2u = -2u=−2So, 'f(u)' is undefined for u=−2u = -2u=−2. Therefore, the range of the function is all real numbers except the value of the function at u=−2u = -2u=−2.

STEP 13

Find the value of the function at u=−2u = -2u=−2.f(−2)=(−2)2+2(−2)+−2+2f(-2) = \frac{(-2)^2 +2(-2) +}{-2 +2}f(−2)=−2+2(−2)2+2(−2)+​

STEP 14

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionDetermine the domain and range of the function $f(u)=\frac{u+1}{1+\frac{1}{u+1}}$ .

Assumptions1. The function is $f(u)=\frac{u+1}{1+\frac{1}{u+1}}$ . We need to find the domain and range of this function

Set the denominator equal to zero and solve for 'u'.
$1+\frac{1}{u+1} =0$

Rearrange the equation to isolate the term with 'u'.
$\frac{1}{u+1} = -1$

olve for 'u' by multiplying both sides by $u+1$ .
$u+1 = -1$

Subtract1 from both sides to solve for 'u'.
$u = -1 -1$

Calculate the value of 'u'.
$u = -2$ So, the function is undefined for $u = -2$ . Therefore, the domain of the function is all real numbers except $-2$ .

implify the function by multiplying the numerator and the denominator by $u+$ .
$f(u) = \frac{(u+)^2}{(u+) +}$

implify the function further.
$f(u) = \frac{u^2 +2u +}{u +2}$

Now, let's find the values that 'f(u)' can take. We can see that 'f(u)' can take all real values except for when the denominator is zero.Set the denominator equal to zero and solve for 'f(u)'.
$u + =0$

Subtract2 from both sides to solve for 'u'.
$u = -2$ So, 'f(u)' is undefined for $u = -2$ . Therefore, the range of the function is all real numbers except the value of the function at $u = -2$ .

Find the value of the function at $u = -2$ .
$f(-2) = \frac{(-2)^2 +2(-2) +}{-2 +2}$