Question
Show that the function $f(x)=4 x^3-18 x^2+27 x-7$ is always increasing on R.
✓
Answer
$f(x)=4 x^3-18 x^2+27 x-7$
$f^{\prime}(x)=12 x^2-36 x+27$
$=3(2 x-3)^2 \geq 0 ; x \in R$
$\therefore f(x)$ is increasing on $R$.
$f^{\prime}(x)=12 x^2-36 x+27$
$=3(2 x-3)^2 \geq 0 ; x \in R$
$\therefore f(x)$ is increasing on $R$.
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