Let f:[a,b]\rightarrow \mathbb{R} be continuous and f(a)

Question

Let   f:[a,b]\rightarrow \mathbb{R}  be continuous and    f(a) < 0 < f(b)  . Define   A=\{x \in [a,b]|f(x) < 0 \}  and consider the proof of the Intermediate Value Theorem. Which of the following claims are True?

Accepted Answer

There exists    c \in (a,b)   such that    f(c) = 0   and   c= \sup A  .

Accepted Answer

If   \sup A=c \in (a,b)  and   f(c) < 0  then there are elements of   A  to the right of   c  and that is a contradiction.

Accepted Answer

If   \sup A=a  then there are elements of is   A  to the right of   a  and that is a contradiction.

Accepted Answer

f(a) < 0 , therefore   a \in A   and   A  is a non-empty subset of   \mathbb{R} .

Accepted Answer

If   A  is a non-empty subset of   \mathbb{R}  then it has a supremum,   \sup A\in \mathbb{R} .

Answer

If   \sup A=c \in (a,b)  and   f(c) > 0  then there are upper bounds of   A  to the left of   c  and that is a contradiction.

Answer

If   \sup A=c \in (a,b)  and   f(c) > 0  then there are elements of   A  to the left of   c  and that is a contradiction.

Answer

If   \sup A=c \in (a,b)  and   f(c) < 0  then there are upper bounds of   A  to the right of   c  and that is a contradiction.

Answer

If   \sup A=a  then there are upper bounds of   A  to the right of   a  and that is a contradiction.

Answer

If   \sup A=b  then there are upper bounds of   A  to the left of   b  and that is a contradiction.

Answer

If   \sup A=b  then there are elements of   A  to the left of   b  and that is a contradiction.

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Let f:[a,b]\rightarrow \mathbb{R} be continuous and f(a) < 0 < f(b) . Def...

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