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

Given the function [tex]f(x)=\frac{x^2 + 8x + 7}{x+1}[/tex], you need to factor the numerator. Find two number whose sum be 8 and whose product be 7. These are 1 and 7, then:

[tex]f(x)=\frac{(x+1)(x+7)}{(x+1)}[/tex]

Then, the denominator is zero when [tex]x=-1[/tex]

Therefore, [tex]x=-1[/tex] does not belong to the Domain of the function. Then, (-1,6) is a discontinuity point.

Simplifying, you get:

[tex]f(x)=x+7[/tex]

You can observe that a linear function is obtained.

This function is equal to zero when [tex]x=-7[/tex], therefore the zero of the function is at (-7,0).

kudzordzifrancis kudzordzifrancis · Answer 2 · 2019-06-08T22:22:56+02:00

ANSWER

Discontinuity:

[tex]x = - 1[/tex]

Zero:

[tex]x = - 7[/tex]

EXPLANATION

The given rational function is

[tex]f(x) = \frac{ {x}^{2} + 8x + 7 }{x + 1} [/tex]

We factor function to obtain:

[tex]f(x) = \frac{(x + 1)(x + 7)}{x + 1} [/tex]

This function is not continuous when

[tex](x + 1) = 0[/tex]

The function is not continuous at

[tex]x = - 1[/tex]

When we simplify the function, we get;

[tex]f(x) = x + 7[/tex]

The zero(s) occur at

[tex]x + 7 = 0[/tex]

[tex]x = - 7[/tex]