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The image of g is not necessarily [-1;1]. However, it contains f(0)-0, which is > 0, and f(1)-1, which is < 0. Per the IVT, any number between these bounds gets reached by g, in particular, 0.
Yes
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ReplyDeleteThe image of g is not necessarily [-1;1]. However, it contains f(0)-0, which is > 0, and f(1)-1, which is < 0. Per the IVT, any number between these bounds gets reached by g, in particular, 0.
ReplyDeleteYes
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