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Question
Show that function f(x) =`3/"x" + 10`, x ≠ 0 is decreasing.
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Solution
f(x) = `3/"x" + 10`
For function to be decreasing, f '(x) < 0
Then f '(x) = `(- 3)/"x"^2 < 0` ....[∵ x ≠ 0, - x2 < 0]
Negative sign indicates that it always decreases as x2 never becomes negative.
Thus, f(x) is a decreasing function for x ≠ 0.
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