So, IVT essentially says if a function $f$ is continuous over an interval $[a,b]$, then the function $f$ will take up all the values between $f(a)$ and $f(b)$ at least once at some point in the interval $[a,b]$.
Now, suppose $f$ is continuous between $[3,7]$ and $f(3) = 4$ and $f(7) = 25$
According to IVT, $f(3) = 4 < m < 25 = f(7)$, i.e., the function takes up all the values between $f(3)$ and $f(7)$ at least once in the interval. But, according to the below picture, clearly, the function within that interval takes up values different that do not satisfy the above inequality. Please help me fill the gap in my understanding of IVT here.
Please do correct me if I'm mistaken.
This doubt arose from the below KhanAcademy question
1 Answer
The intermediate value theorem tells you that, in that context, for each $y\in[4,25]$, there is some $x\in[3,4]$ such that $f(x)=y$. It does not tell you that if $x\in[3,4]$, then $f(x)\in[4,25]$. So, there is no contradiction if $f\bigl([3,4]\bigr)$ contains values outside $[4,25]$.
