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Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using interval notation. If the answer cannot be expressed as an interval, enter EMPTY or ∅.)

f(x) = (5 − x^2) / x

my work:
a) f'(x)=(-x^2-5)/x^2

b) set numerator equal to zero to find critical points --> c = +/- square root of 5

c) do a number line of x-values separated by the critical points to determine increasing and decreasing --> find increasing (-square root 5, square root 5); decreasing (- infinity, -square root 5), (square root 5, infinity)

This is incorrect. Where did I go wrong?

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3 answers

  1. look at the slope in the interval, it is negative, decreasin from -sqrt5 to sqrt5

    then look at x=100, slope is negative again.
    and at x=-100, negative again.

    Look at the graph here: http://www.wolframalpha.com/input/?i=(5+%E2%88%92+x%5E2)+%2F+x

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  2. f'(x) = ( - x² - 5 ) / x² = - 1 - 5 / x² = - ( 1 + 5 / x² )

    function is increasing on the interval, where f′(x) > 0

    function is decreasing on the interval, where f′(x) < 0

    In this case:

    f'(x) = - ( 1 + 5 / x² ) is always < 0

    becouse:

    when x-> - ∞

    then

    f'(x) = - ( 1 + 5 / x² ) -> - 1

    when x-> 0

    then

    f'(x) = - ( 1 + 5 / x² ) -> - ∞

    when x-> ∞

    then

    f'(x) = - ( 1 + 5 / x² ) -> - 1

    So f'(x) is always < 0

    Your function decreasing on the interval (- ∞ ,∞ )

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  3. Thank you so much! I understand now.

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