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Question
Solve the following:
Find the intervals on which the function f(x) = `x/logx` is increasing and decreasing.
Solution
f(x) = `x/logx`
∴ f'(x) = `d/dx(x/logx)`
= `((logx).d/dx(x) - x*d/dx(logx))/(logx)^2`
= `((logx) xx 1 - x xx 1/x)/(logx)^2`
= `(logx - 1)/(log x)^2`
f is increasing if f'(x) ≥ 0
i.e. if `(log x - 1)/(logx)^2 ≥ 0`
i.e. if log x – 1 ≥ 0 ...[∵ (log x)2 > 0]
i.e. if log x ≥ 1
i.e. if log x ≥ log e ...[∵ log e = 1]
i.e. if x ≥ e
∴ f is increasing on `[e, oo)`
f is decreasing if f'(x) ≤ 0
i.e. if `(log x - 1)/(logx)^2 ≤ 0`
i.e. if log x – 1 ≤ 0 ...[∵ (log x)2 > 0]
i.e. if log x ≤ 1
i.e. if log x ≤ log e
i.e. if x ≤ e
Also, x > 0 and x ≠ 1 because f(x) = `x/logx` is not defined at x = 1
∴ f is decreasing in (0, e] – {1}
Hence, f is increasing in `[e, oo)` and decreasing in (0, e] – {1}.
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