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Mathematics
Calculus
find max and min of e^x-e^(3x) on interval where x is greater than or equal to 0 and less than or equal to 1
1 answer
Since e^3x > e^x when x > 0,
min at x=0: e^0 - e^0 = 0
max at x=1: e^3 - e^1
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