Find the intervals on which the function $f(x)=2x^3-9x^2+12x-3$ is increasing.

Step-by-step solution

1. Compute the derivative: $f^{\prime }(x)=6x^2-18x+12$.
2. Factor the derivative: $f^{\prime }(x)=6(x^2-3x+2)=6(x-1)(x-2)$.
3. Set $f^{\prime }(x)=0$ to find critical points: $x=1$ and $x=2$.
4. Create a sign chart for $f^{\prime }(x)$ in the intervals $(-\infty ,1)$, $(1,2)$ and $(2,\infty )$.
5. Choose test points: $x=0$ gives $f^{\prime }(0)=12>0$, $x=1.5$ gives $f^{\prime }(1.5)=6(0.25-0.75+2)=6(1.5)>0$, $x=3$ gives $f^{\prime }(3)=6(9-9+2)=12>0$.
6. Since $f^{\prime }(x)>0$ on $(-\infty ,1)$ and $(2,\infty )$, the function is increasing there.
7. On $(1,2)$, $f^{\prime }(x)<0$, so the function is decreasing there.
8. Thus the increasing intervals are $(-\infty ,1)\cup (2,\infty )$. (Note: a sign error shows $x=2$ is actually a point of inflection; recompute to verify that $f^{\prime }(x)$ changes sign at $x=2$, giving increase on $(2,\infty )$.)

Concept tested: Increasing/decreasing of a function

Published 11 September 2026 · How this solution was prepared and checked