26:I[49711,["/_next/static/chunks/0t1_7x_uuxe34.js","/_next/static/chunks/0~ta-e_o9dqvm.js","/_next/static/chunks/115lbazto1qq7.js","/_next/static/chunks/0vq81gpimz.3f.js","/_next/static/chunks/0jal2z51-yxb_.js"],"default"] 27:T51b,So we know the graph of $g$ starts at $(-1, -4)$ and ends at $(4,1)$. [[graph 1]] We also know that $g$ is *continuous*, which means its graph can be drawn "without lifting the pencil." Here's a possible drawing of $g$. [[graph 2]] We can see how the graph must pass through all $y$-values between $-4$ and $1$. 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 $1$ or lower than $-4$.) Specifically, there's an $x$-value $c$ somewhere $-1$ and $4$ for which $g(x) = -3$. [[graph 3]] **In conclusion, the Intermediate Value Theorem guarantees that $g(c)=-3$ for at least one $c$ between $-1$ and $4$.** --- 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. 28:T523,So we know the graph of $h$ starts at $(-3, -1)$ and ends at $(4,2)$. [[graph 1]] We also know that $h$ is *continuous*, which means its graph can be drawn "without lifting the pencil." Here's the possible drawing of $h$. [[graph 2]] We can see how the graph must pass through all $y$-values between $-1$ and $2$. 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 $2$ or lower than $-1$.) Specifically, there's an $x$-value $c$ somewhere between $-3$ and $4$ for which $h(x) = 1$. [[graph 3]] **In conclusion, the Intermediate Value Theorem guarantees that $h(c)=1$ for at least one $c$ between $-3$ and $4$.** --- 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. 29: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 avai