Question

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

Answer 1

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