Intermediate Value Theorem guarantee

Question

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?

Answer 1

Since $f(-3)=-2<0<3=f(6)$, 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 $x\in[-3,6]$ such that $f(x)=c$ for any $c\in[-2,3]$. 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 $x\in(-3,6)$. I hope this helps make the solution a little bit more clear.

Answer 2

It means that for every value $\,c\in [f(-3),f(6)]=[-2,3]\,$ there exists at least one value $\,x_c\in [-3,6]\,$ s.t. $\,f(x_c)=c\,$.

The above, for example, tells us the function has a zero in $\,[-3,6]\,$...

Answer 3

The actual statement of the Intermediate Value Theorem is as follows.

Let $f$ be a continuous function on the closed interval $[a,b].$ Then for any $c \in (f(a),f(b))$ (this is an open interval), there exists $x \in (a,b)$ such that $f(x) = c.$

In your case, it guarantees the existence of a zero in the interval $[-3,6].$

Something interesting to note is that if $f$ is differentiable on $[a,b],$ then $f^{\prime}$ has the intermediate value property. This is known as Darboux's Theorem and is a nice exercise.