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Maths help ?

I tried to answer this q by subbing in (-2) , and (-1) but still go the wrong answer could someone please explain 

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  • Pope
    Lv 7
    2 months ago

    You must be referring to part (a). What did you get when you evaluated the function at -2 and -1? Is one negative and the other positive? The function is a polynomial, hence continuous an all teal numbers. It cannot move between negative and positive without meeting zero somewhere on the way.

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  • ted s
    Lv 7
    2 months ago

    a) f(-2) > 0 while f(-1) < 0 and f is continuous ====> it crosses the x axis

    c)f ' ( x) = 4 ( x³ - 1)====> x = 1 is a critical value

    b)...divide f(x) by ' x - 2 '  to find a ,  b , c.....not hard

    d) 4th degree means starts high and ends high...using parts b & c we can deduce that a 2nd root exists...function decreases for x < 1 and increases for x > 1

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  • Anonymous
    2 months ago

    f(x) = x^4 - 4x - 8 = 16 for x = -2.  the function = -3 for x = -1.  Since you go from 16 to -3, the function must equal 0 at some point between those two numbers and that is a root.  Specifically, about x = -1.296

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