Question
Show that the function $f$ given by $f(x) = x^3 – 3x^2 + 4x, x \in R$ is increasing on $R$.
✓
Answer
Not that
$f ′(x) = 3x^2 – 6x + 4$
$= 3(x^2 – 2x + 1) + 1$
$= 3(x – 1)^2 + 1 > 0$, in every interval of R
Therefore, the function f is increasing on R.
$f ′(x) = 3x^2 – 6x + 4$
$= 3(x^2 – 2x + 1) + 1$
$= 3(x – 1)^2 + 1 > 0$, in every interval of R
Therefore, the function f is increasing on R.
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