Question

Question 1(Multiple Choice Worth 5 points) (04.02 MC) What are the domain and range of the function f of x is equal to the quantity x squared plus 8x plus 7 end quantity divided by the quantity x plus 1 end quantity? D: {x ∊ ℝ | x ≠ 1}; R: {y ∊ ℝ | y ≠ 0} D: {x ∊ ℝ | x ≠ −1}; R: {y ∊ ℝ | y ≠ −6} D: {x ∊ ℝ | x ≠ −7}; R: {y ∊ ℝ | y ≠ 0} D: {x ∊ ℝ | x ≠ −1}; R: {y ∊ ℝ | y ≠ 6}

Answer 1

The domain and the range of the function is (d) D: {x ∊ ℝ | x ≠ -1}; R: {y ∊ ℝ | y ≠ 6}

The complete question

What are the domain and range of the function f(x) = x^2 + 8x + 7/x + 1

The domain of the function

We have the function to be

$f(x) = \frac{x^2 + 8x + 7}{x + 1}$

Set the denominator to 0

x + 1 = 0

Solve for x

x = -1

This means that the domain of the function is the set of all real numbers except -1

Hence, the domain of the function is {x ∊ ℝ | x ≠ -1}

The range of the function

We have the function to be

$f(x) = \frac{x^2 + 8x + 7}{x + 1}$

Factorize the numerator

$f(x) = \frac{(x + 1)(x + 7)}{x + 1}$

Simplify

f(x) = x + 7

The function is undefined at x= -1.

So, we have:

f(-1) = -1 + 7

f(-1) = 6

This means that the range of the function is the set of all real numbers except 6

Hence, the range of the function is {y ∊ ℝ | y ≠ 6}

Read more about domain and range at:

https://brainly.com/question/10197594

#SPJ1