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Question
Show that the function `f(x) = x^3 - 3x^2 + 6x - 100` is increasing on R
Solution 1
`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.
Solution 2
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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