Let $f(x)=x^3+x+33$, which is continuous everywhere. On which interval does the Intermediate Value Theorem guarantee a solution of $f(x)=0$?

A continuous function $g$ has $g(3)=-5$ and $g(7)=11$. Which value is the Intermediate Value Theorem guaranteed to take somewhere on $\left(3,7\right)$?

A function $f$ is continuous on $\left[a,b\right]$, and $k$ lies strictly between $f(a)$ and $f(b)$. The Intermediate Value Theorem guarantees a value $c$ with $f(c)=k$. Where must that $c$ lie?

Let $f(x)=\frac{1}{x}$ on $\left[-4,4\right]$, with $f(-4)=-\frac{1}{4}$ and $f(4)=\frac{1}{4}$. Does the Intermediate Value Theorem guarantee a value $c$ in $\left(-4,4\right)$ with $f(c)=0$?

A continuous function $r$ has $r(2)=-4$ and $r(10)=1$. Which value is the Intermediate Value Theorem guaranteed to take somewhere on $\left(2,10\right)$?

A continuous function $f$ on $\left[a,b\right]$ has $f(a)<k<f(b)$. What does the Intermediate Value Theorem conclude?

A student writes: since $f(1)=-5$ and $f(6)=8$, the Intermediate Value Theorem guarantees exactly one value $c$ in $\left(1,6\right)$ with $f(c)=0$. What is wrong with the justification?

A student writes: since $s(0)=-4$ and $s(6)=2$, the Intermediate Value Theorem guarantees exactly one value $c$ in $\left(0,6\right)$ with $s(c)=0$. What is wrong with the justification?

A continuous function $f$ has $f(6)=-2$ and $f(12)=12$. Which value is the Intermediate Value Theorem guaranteed to take somewhere on $\left(6,12\right)$?

Let $f(x)=x^3+x+14$, which is continuous everywhere. On which interval does the Intermediate Value Theorem guarantee a solution of $f(x)=0$?

Let $f(x)=x^3+x+8$, which is continuous everywhere. On which interval does the Intermediate Value Theorem guarantee a solution of $f(x)=0$?

Let $f(x)=\frac{1}{x}$ on $\left[-8,8\right]$, with $f(-8)=-\frac{1}{8}$ and $f(8)=\frac{1}{8}$. Does the Intermediate Value Theorem guarantee a value $c$ in $\left(-8,8\right)$ with $f(c)=0$?

A continuous function $q$ has $q(5)=-3$ and $q(12)=8$. Which value is the Intermediate Value Theorem guaranteed to take somewhere on $\left(5,12\right)$?

A student writes: since $r(2)=-5$ and $r(6)=2$, the Intermediate Value Theorem guarantees exactly one value $c$ in $\left(2,6\right)$ with $r(c)=0$. What is wrong with the justification?

For which target value $k$ does the Intermediate Value Theorem say nothing, even for a continuous $f$ on $\left[a,b\right]$?

Let $f(x)=\frac{1}{x}$ on $\left[-3,3\right]$, with $f(-3)=-\frac{1}{3}$ and $f(3)=\frac{1}{3}$. Does the Intermediate Value Theorem guarantee a value $c$ in $\left(-3,3\right)$ with $f(c)=0$?

Let $f(x)=x^3+x-4$, which is continuous everywhere. On which interval does the Intermediate Value Theorem guarantee a solution of $f(x)=0$?

A student writes: since $q(4)=-1$ and $q(6)=5$, the Intermediate Value Theorem guarantees exactly one value $c$ in $\left(4,6\right)$ with $q(c)=0$. What is wrong with the justification?

The Intermediate Value Theorem applies to a function $f$ on $\left[a,b\right]$ under which condition?

Let $f(x)=x^3+x-27$, which is continuous everywhere. On which interval does the Intermediate Value Theorem guarantee a solution of $f(x)=0$?