28:T544,So we know the graph of $f$ starts at $(-5,0)$ and ends at $(0,5)$. [[graph 1]] We also know that $f$ is *continuous*, which means its graph can be drawn "without lifting the pencil." Here's a possible drawing of $f$. [[graph 2]] We can see how the graph must pass through all $y$-values between $0$ and $5$. This is basically the meaning of the Intermediate Value Theorem! (Notice that the graph doesn't necessarily have to pass through $y$-values higher than $5$ or lower than $0$.) Specifically, there's an $x$-value $c$ somewhere between $-5$ and $0$ for which $f(x) = 2$. [[graph 3]] **In conclusion, the Intermediate Value Theorem guarantees that $f(c) = 2$ for at least one $c$ between $-5$ and $0$.** --- Licensing information: The question comes from Khan Academy, and is licensed under a creative commons license. The original exercise containing this question can be found here: Using the intermediate value theorem. Note: All Khan Academy content is available for free at www.khanacademy.org.29:T533,The intermediate value theorem has two conditions: 1. Function $g$ is continuous over the interval $[a,b]$. 2. The value $d$ lies between $g(a)$ and $g(b)$. We *must* establish these two conditions to conclude that there is a value $c$ in the interval $[a,b]$ for which $g(c) = d$ (another way to phrase this conclusion is that the equation $g(x)=d$ has a solution where $a \leq x \leq b$). Did Min establish these two conditions? Min did establish that $-50$ is between $g(12)$ and $g(21)$. However, he didn't establish that $g$ is continuous. To do that, Min could mention that we were given that $g$ is continuous in his answer. **In conclusion, Min's justification is incomplete. He should've mentioned that $g$ is continuous.** --- Licensing information: The question comes from Khan Academy, and is licensed under a creative commons license. The original quiz containing this question can be found here: Quiz 6. Note: All Khan Academy content is available for free at www.khanacademy.org.