Greatest and Least Values on a Closed Interval

On a closed interval, the greatest or least value can occur at a stationary point or at an endpoint. So find the stationary points inside the interval, then compare their $y$-values with the values at both ends.

Find the greatest and least values of $y = x^3 - 3x + 2$ for $-2 \le x \le 2$. [6]

The working

Step 1, find the stationary points: $\frac{dy}{dx} = 3x^2 - 3 = 3(x^2 - 1) = 0 \implies x = -1 \ \text{or} \ x = 1 \quad \text{(M1, A1)}$

Both lie inside $[-2, 2]$, so both count.

Step 2, evaluate $y$ at the stationary points:

• $x = -1$: $\ y = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4$
• $x = 1$: $\ y = 1 - 3 + 2 = 0$ (M1)

Step 3, evaluate $y$ at the endpoints:

• $x = -2$: $\ y = -8 + 6 + 2 = 0$
• $x = 2$: $\ y = 8 - 6 + 2 = 4$ (M1)

Step 4, compare all four values $\{4, 0, 0, 4\}$: $\text{greatest value} = 4, \qquad \text{least value} = 0. \quad \text{(A1, A1)}$

Where the marks are won and lost

• Check the endpoints, not just the turning points. Here the endpoint $x = 2$ ties for the greatest value.
• Only stationary points inside the interval count; discard any that fall outside $[-2, 2]$.
• Compare the actual $y$-values, then state which is greatest and which least.

Common mistakes

• Reporting only the turning-point values and missing that an endpoint is just as high.
• Giving the $x$-values instead of the $y$-values as the “greatest/least value”.
• Including a stationary point that lies outside the interval.

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