glycleWogry

2022-06-22

I have a problem with the Intermediate value theorem. For example if I have the function $(x)=-4{x}^{2}+12x$, I can get for example all the values from $x=0$ to $x=2$, so $f(0)=0$ and $f(2)=8$, 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.

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

grcalia1

Beginner2022-06-23Added 23 answers

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

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.

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.

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