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प्रश्न
Solve the following : Find the intervals on which the function y = xx, (x > 0) is increasing and decreasing.
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उत्तर
y = xx
∴ log y = log xx = x log x
Differentiating both sides w.r.t. x, we get
`(1)/y.dy/dx = d/dx(x log x)`
= `x.d/dx(log x) + (log x).d/dx(x)`
= `x xx (1)/x + (log x) xx 1`
∴ `dy/dx = y(1 + logx)`
= xx(1 + log x)
y is increasing if `dy/dx ≥ 0`
i.e. if xx (1 + log x) ≥ 0
i.e. if 1 + log x ≥ 0 ...[∵ x > 0]
i.e. if log x ≥ – 1
i.e. if log x ≥ – log e ...[∵ log e = 1]
i.e. if log x ≥ log `(1)/e`
i.e. if x ≥ `(1)/e`
∴ y is increasing in `[1/e, oo)`
y is decreasing if `dy/dx ≤ 0`
i.e. if xx (1 + log x) ≤ 0
i.e. if 1 + log x ≤ 0 ...[∵ x > 0]
i.e. if log x ≤ – 1
i.e. if log x ≤ – log e ...[∵ log e = 1]
i.e. if log x ≤ log `(1)/e`
i.e. if x ≤ `(1)/e`, where x > 0
∴ y is decreasing is `(0, 1/e]`
Hence, the given function is increasing `[1/e, oo)`
and decreasing in `(0, 1/e]`.
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