How would I solve this?
The maximum was straight forward, x = 1. But I can't find the minimum within the interval using derivatives.
Any help would be appreciated. Thanks in advance.
2026-04-07 18:09:02.1775585342
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global minimum for $2x^3+3x^2-12x+1$ on intervals between 0 and 2
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A local minimum means the graph is decreasing, then turns around and starts increasing. That's where the slope is zero. If there is no such valley in the interval in question, then the minimum must be at one of the boundaries of the interval. So check which boundary is lower and you've found the minimum.
HINT: If there is no local minimum on your interval simply check the endpoints.