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Question
Show that the function `f(x) = x/3 + 3/x` decreases in the intervals (–3, 0) ∪ (0, 3).
Sum
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Solution
`f(x) = x/3 + 3/x`
f'(x) = `1/3 - 3/x^2`
f'(x) = 0
`1/3 - 3/x^2 = 0`
`(x^2 - 9)/(3x^2) = 0`
x2 – 9 = 0, x not equal sig 0
(x + 3)(x – 3) = 0
x = –3, 3

Here x ≠ 0 the function decreases in the interval (–3, 0) ∪ (0, 3).
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