Montenovofe

2022-07-10

I'm doing a review packet for Calculus and I'm not really sure what it is asking for the answer?
The question is: Let f be a continuous function on the closed interval [-3, 6]. If f(-3)=-2 and f(6)=3, what does the Intermediate Value Theorem guarantee? I get that the intermediate value theorem basically means but not really sure how to explain it?

Jamiya Costa

Expert

Since $f\left(-3\right)=-2<0<3=f\left(6\right)$, we can guarantee that the function has a zero in the interval [−3,6]. We cannot conclude it has only one, though (it may be many zeros).

EDIT: As has already been pointed out elsewhere, the IVT guarantees the existence of at least one $x\in \left[-3,6\right]$ such that $f\left(x\right)=c$ for any $c\in \left[-2,3\right]$. Note that the fact that there is a zero may be important (for example, you couldn't define a rational function over this domain with this particular function in the denominator), or you may be more interested in the fact that it attains the value y=1 for some $x\in \left(-3,6\right)$. I hope this helps make the solution a little bit more clear.

bandikizaui

Expert

It means that for every value $\phantom{\rule{thinmathspace}{0ex}}c\in \left[f\left(-3\right),f\left(6\right)\right]=\left[-2,3\right]\phantom{\rule{thinmathspace}{0ex}}$ there exists at least one value $\phantom{\rule{thinmathspace}{0ex}}{x}_{c}\in \left[-3,6\right]\phantom{\rule{thinmathspace}{0ex}}$ s.t. $\phantom{\rule{thinmathspace}{0ex}}f\left({x}_{c}\right)=c\phantom{\rule{thinmathspace}{0ex}}$.

The above, for example, tells us the function has a zero in $\phantom{\rule{thinmathspace}{0ex}}\left[-3,6\right]\phantom{\rule{thinmathspace}{0ex}}$...

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