Verify MVT hypotheses and find the guaranteed value of c

Since $f(x) = x^3 - 3x + 2$ is a polynomial, it is continuous on $[0,2]$ and differentiable on $(0,2)$, so the Mean Value Theorem applies. We have $f(2) = 8 - 6 + 2 = 4$ and $f(0) = 2$. The slope of the secant line is $\frac{f(2)-f(0)}{2-0} = \frac{4 - 2}{2} = 1$.

Now compute $f'(x) = 3x^2 - 3$ and set it equal to the secant slope: $3c^2 - 3 = 1$, so $3c^2 = 4$, giving $c^2 = \frac{4}{3}$, thus $c = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}$. Only $c = \frac{2\sqrt{3}}{3} \approx 1.1547$ lies in $(0,2)$, so this is the guaranteed value.

At $c = \frac{2\sqrt{3}}{3}$, the derivative $f'(c) = 1$, which is exactly equal to the slope of the secant line joining $(0,2)$ and $(2,4)$, computed earlier as $1$.

Since the tangent line at $x=c$ and the secant line through the endpoints have the same slope, the two lines are parallel — this is the geometric meaning of the MVT: somewhere on the curve, the tangent is parallel to the secant.

The guaranteed value of c is positive. It was found to be two divided by the square root of three, approximately one point one five, which is a positive number lying inside the interval from zero to two. The other algebraic solution was negative and was rejected because it fell outside the open interval.