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

Question

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

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Josephine 28 mins 2021-10-13T06:17:07+00:00 2 Answers 0

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    0
    2021-10-13T06:18:23+00:00

    Answer:

    – Discontinuity at (-1,6)

     – The zero is at (-7,0)

    Step-by-step explanation:

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

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

    Then, the denominator is zero when x=-1

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

    Simplifying, you get:

    f(x)=x+7

    You can observe that a linear function is obtained.

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

    0
    2021-10-13T06:18:58+00:00

    ANSWER

    Discontinuity:

    x =  - 1

    Zero:

    x =  - 7

    EXPLANATION

    The given rational function is

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

    We factor function to obtain:

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

    This function is not continuous when

    (x + 1) = 0

    The function is not continuous at

    x =  - 1

    When we simplify the function, we get;

    f(x) = x + 7

    The zero(s) occur at

    x + 7 = 0

    x =  - 7

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