Question
Prove that the following function are increasing on R.
$f(x) = 3x^5 + 40x^3 + 240x$

Answer

$f(x) = 3x^5 + 40x^3 + 240x$
$f'(x) = 15x^4 + 120x^2 + 240$
$= 15(x^4 + 8x^2 +16)$
$=15(\text{x}^2+4)^2>0,\forall\ \text{x}\in\text{R}$ $[\because\ 15>0\text{ and }(\text{x}^2+4)^2>0]$
So, $f(x)$ increasing on $R$.

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