Question
Prove that the following function are increasing on R.
f(x) = 3x5 + 40x3 + 240x

Answer

f(x) = 3x5 + 40x3 + 240x
f'(x) = 15x4 + 120x2 + 240
= 15(x4 + 8x2 +16)
$=15(\text{x}^2+4)^2>0,\forall\ \text{x}\in\text{R}$ $[\because\ 15>0\text{ and }(\text{x}^2+4)^2>0]$
So, f(x) increasing on R.

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