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Show that the Function `F(X) = Xcuberoot3 - 3xsqrt2 + 6x - 100` is Increasing on R

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प्रश्न

Show that the function `f(x) = x^3 - 3x^2 + 6x - 100` is increasing on R

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उत्तर १

`f(x) = x^3 - 3x^2 + 6x - 100`

`f'(x) = 3x^2 - 6x + 6`

`= 3(x^2 -  2x + 1 ) + 3`

=`3(x+1)^2 + 3 > 0`

For all values of x, `(x - 1)^2` is always positve

`:. f'(x) > 0`

So, f (x) is increasing function.

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उत्तर २

The given function is

f(x) = x3 − 3x2 + 6x −100

∴f'(x) = 3x2 − 6x + 6

=3(x2 − 2x +2)

=3(x2 − 2x + 1) + 3

=3(x−1)2+3

For f(x) to be increasing, we must have f'(x) > 0

Now, 3(x−1)2 ≥ 0  ∀x ∈ R

⇒ 3(x − 1)2 + 3 > 0  ∀x ∈ R

⇒ f'(x) > 0    ∀x ∈ R

Hence, the given function is increasing on R

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2016-2017 (March) All India Set 1

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