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Solve the following : Find the intervals on which the function y = xx, (x > 0) is increasing and decreasing. - Mathematics and Statistics

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प्रश्न

Solve the following : Find the intervals on which the function y = xx, (x > 0) is increasing and decreasing.

योग

उत्तर

y = x
∴ 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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अध्याय 2: Applications of Derivatives - Miscellaneous Exercise 2 [पृष्ठ ९३]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
अध्याय 2 Applications of Derivatives
Miscellaneous Exercise 2 | Q 10 | पृष्ठ ९३

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