A vanishing derivative at an endpoint of the interval
A student is analysing f(x) = x^3 - 3x on the closed interval [0, 2]. They compute f'(x) = 3x^2 - 3, solve f'(x) = 0 to get x = 1 (rejecting x = -1 as outside the interval), and then evaluate f(1) = -2. They conclude that the maximum value of f on [0, 2] is -2 and that f has no maximum because the only critical point gives a minimum. They also note f'(0) = -3 and f'(2) = 9, so the function is 'increasing at the right end'. A classmate claims that because f'(2) > 0, the function must still be rising when it leaves the interval, so the largest value must occur 'just past' x = 2.
1. Identify the error in the student's conclusion about the maximum on [0, 2], and give the correct maximum and minimum values with justification.
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The student has forgotten that a continuous function on a closed interval attains its extrema either at interior critical points or at the endpoints. The candidate set is {x = 0, x = 1, x = 2}.
- f(0) = 0
- f(1) = 1 - 3 = -2
- f(2) = 8 - 6 = 2
So the maximum value is 2, attained at x = 2, and the minimum is -2, attained at x = 1.
The critical point x = 1 is indeed a local minimum, but it is not the global minimum story on its own - it happens to be, but only after checking endpoints. The maximum being at an endpoint is entirely normal: f is increasing on [1, 2] since f'(x) = 3(x^2 - 1) > 0 there, so the largest value is at the right endpoint.
2. Explain why the classmate's reasoning about f'(2) > 0 implying a larger value 'just past' x = 2 is invalid, and state what f'(2) > 0 actually tells you.
Read the worked answer
The domain is restricted to [0, 2]; points 'just past' x = 2 are not in the domain, so they cannot compete for the maximum. The sign of f'(2) tells you about the local behaviour of f near x = 2 on the left: since f'(2) = 9 > 0, f is increasing as it approaches x = 2 from within the interval, which is exactly why f(2) is the largest value on [0, 2].
A derivative at an endpoint is a one-sided statement about the trend inside the interval. It says nothing about values outside the domain. The classmate has confused 'the function is still rising at the boundary' with 'the function continues to rise beyond the boundary' - the second claim requires information about f outside [0, 2], which is not given.
3. Suppose the interval is changed to the open interval (0, 2). Describe precisely how the answer changes and why.
Read the worked answer
On the open interval (0, 2) the endpoints are excluded, so f(0) = 0 and f(2) = 2 are not attained. The function still has its local minimum -2 at x = 1 (which is inside), but it has no maximum: for any point x close to 2 from below, f(x) < 2, yet values approach 2 as x → 2⁻. So the supremum is 2 but no maximum exists.
This shows the closed-interval hypothesis in the extreme value theorem is doing real work. Removing even one endpoint can destroy attainment of an extremum, even though the supremum is unchanged.