हिंदी

Show that f(x) = 3x+13x is increasing in (13,1) and decreasing in (19,13).

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प्रश्न

show that f(x) = `3x + (1)/(3x)` is increasing in `(1/3, 1)` and decreasing in `(1/9, 1/3)`.

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उत्तर

f(x) = `3x + (1)/(3x)`

∴ f'(x) = `3d/dx(x) + (1)/(3)d/dx(x^-1)`

= `3 xx 1 + (1)/(3)(-1) x^-2`

= `3 - (1)/(3x^2)`
Now, f is increasing if f'(x) > 0 and is decreasing if f'(x) < 0.

 Let `x ∈ (1/3, 3)`.

Then `(1)/(3) < x < 1`

∴ `(1)/(9) < x^2 < 1`

∴ `(1)/(3) < 3x^2 < 3`

∴ `3 >(1)/(3x^2) > (1)/(3)`

∴ `-3 < - (1)/(3x^2) < - (1)/(3)`

∴ `3 - 3 < 3 - (1)/(3x^2) < 3 - (1)/(3)`

∴ `0 < f'(x) < (8)/(3)`

∴ f'(x) > 0 for all x ∈ `(1/3, 1)`

∴ f is increasing in rhe interval `(1/3, 1)`

Let x ∈ `(1/9, 1/3)`.

Then `(1)/(9) < x < (1)/(3)`

∴ `(1)/(81) < x^2  < (1)/(9)`

∴ `(1)/(27) < 3x^2 < (1)/(3)`

∴ `27 > (1)/(3x^2) > 3`

∴ `-27 < -(1)/(3x^2) < - 3`

∴ `3 - 27 < 3 - (1)/(3x^2) < 3 - 3`

∴ – 24 < f'(x) < 0

∴ f'(x) < 0 for all x ∈ `(1/9, 1/3)`

∴ f is decreasing in the interval `(1/9, 1/3)`.

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अध्याय 2: Applications of Derivatives - Exercise 2.4 [पृष्ठ ९०]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 2 Applications of Derivatives
Exercise 2.4 | Q 7 | पृष्ठ ९०

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