Q8 of 39 Page 1

Show that the function f(x) = 4x3 − 18x2 + 27x − 7 is always increasing on IR.

Given: f(x) = 4x3 − 18x2 + 27x − 7.


Differentiating both sides w.r.t x, we get,


f’(x) = 12 x2 − 36x + 27


⇒ f’(x) = 3(4x2 − 12x + 9)


⇒ f’(x) = 3(2x − 3)2


∴ f’(x) is positive for all the real values of x.


∴ f(x) is increasing on R.


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