glycleWogry

2022-06-22

I have a problem with the Intermediate value theorem. For example if I have the function $\left(x\right)=-4{x}^{2}+12x$, I can get for example all the values from $x=0$ to $x=2$, so $f\left(0\right)=0$ and $f\left(2\right)=8$, with the Intermediate value theorem, I know that the function takes all the values from 0 to 8. But it also takes the value 9 when x goes from 0 to 2, so with the Intermediate value theorem I can´t know all the value that the function takes.

Can anyone explain me why this happens in this example, and obviously in other example.

grcalia1

The IVT claims that if $f:\left[a,b\right]\to \mathbb{R}$ is continuous then $f$ has to take on all value between $f\left(a\right)$ and $f\left(b\right)$ for at least one $x\in \left[a,b\right]$.

In your example, the function achieves a extremum inside the interval, such a situation is not covered by IVT. The IVT has a limited set of assumptions, and knowing values on the boundary with guaranteed continuity is not enough to characterize all values inside that the function will take.

Do you have a similar question?