interval [ - 2 , 3 ]
f(x) = x^3 + 4x^2 - 1
f(-2) = (-2)^3 + 4(-2)^2 -1
f(-2) = - 8 + 16 -1
f(-2) = 7
f(3) = (3)^3 + 4(3)^2 - 1
f(3) = 62
as both values f(-2) and f(3) are positive
so,
the intermediate theorem does not guarantee there is a zero in the interval [ - 2 , 3 ]
5 pts) Does the Intermediate Value Theorem guarantee there is a zero in the interval [-2,...
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