Let $g(x)=x^3-3x^2+2x$ defined on the interval $[0,2]$. Verify the hypotheses of Rolle’s theorem and find all $c\in (0,2)$ such that $g^{\prime }(c)=0$.

Step-by-step solution

1. Step 1: Check continuity on $[0,2]$ – $g(x)$ is a polynomial, hence continuous.
2. Step 2: Check differentiability on $(0,2)$ – polynomials are differentiable everywhere, so $g^{\prime }(x)$ exists on $(0,2)$.
3. Step 3: Compute $g(0)=0$ and $g(2)=8-12+4=0$; thus $g(0)=g(2)$.
4. Step 4: By Rolle’s theorem, there exists at least one $c$ in $(0,2)$ with $g^{\prime }(c)=0$.
5. Step 5: Find $g^{\prime }(x)=3x^2-6x+2$ and solve $3x^2-6x+2=0$.
6. Step 6: Use quadratic formula: $x=\frac{6\pm \sqrt{36-24}}{6}=1\pm \frac{\sqrt{3}}{3}$.
7. Step 7: Verify that both solutions lie in $(0,2)$; indeed $1-\frac{\sqrt{3}}{3}\approx 0.423$ and $1+\frac{\sqrt{3}}{3}\approx 1.577$.
8. Step 8: State the two values of $c$ as the required points.

Concept tested: Rolle’s theorem

Published 13 October 2026 · How this solution was prepared and checked