 # The Intermediate Value Theorem states that, for a continuous function f : [ a , b ddcon4r 2022-07-10 Answered
The Intermediate Value Theorem states that, for a continuous function $f:\left[a,b\right]\to \mathbb{R}$, if $f\left(a\right), then there exists a $c\in \left(a,b\right)$ such that $f\left(c\right)=d.$

I wonder if I change the hypothesis of $f\left(a\right) to $f\left(a\right)>d>f\left(b\right)$, the result still holds. I believe so, since $f$ assumes a fixed point $f\left(x\right)=x$ in $\left[a,b\right]$, so we would have $c=d$, although I'm not completely sure.

I need this result in order to prove the set $X=\left\{x\in \left[a,b\right]\phantom{\rule{thinmathspace}{0ex}}s.t.\phantom{\rule{thinmathspace}{0ex}}f|\left[a,x\right]\phantom{\rule{thinmathspace}{0ex}}\text{is bounded}\right\}$, with a continuous $f$, is not empty.
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I don't know how you're getting that $f$ has a fixed point, or how that would imply that $c=d$. You seem to be confusing inputs and outputs of $f$: we know nothing at all about how $f\left(a\right)$ and $f\left(b\right)$ are related to $a$ and $b$.

However, you can instead just apply the Intermediate Value Theorem to the function $g\left(x\right)=-f\left(x\right)$, which satisfies $g\left(a\right)<-d, to get some $c$ such that $g\left(c\right)=-d$, so $f\left(c\right)=d$.

(It is not clear to me what your question has to do with proving that the set $X$ is nonempty, though.)

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