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