The intermediate value theorem is used to establish that a function passes through a certain yvalue and relies heavily on continuity. First, let's look at the theorem itself.
Intermediate Value Theorem 

For a continuous function, \(f(x)\) on an interval \([a,b]\), if \( t \) is between \(f(a)\) and \(f(b)\), 
Notice that the theorem just tells you that the value \(x=c\) exists but doesn't tell you what it is or how to find it.
To get the idea of this theorem clear in your head, here are some great videos for you to watch. They use graphs to help you understand what the theorem means.
Here is a video that shows, graphically, how the intermediate value theorem works. She uses color in her graph to make it easy to follow.
video by Krista King Math 

Here is a great video that clearly explains the intermediate value theorem more from a mathematical point of view than in the previous video.
video by PatrickJMT 

Application of the Intermediate Value Theorem   Here is a great video showing a nonstandard application of the IVT. To work this problem, he uses the definition of the limit. Don't skip this video. It will help you understand limits, continuity and the IVT.
video by Dr Chris Tisdell 

Use the intermediate value theorem to solve these problems. Give your answers in exact form.
Show that \(x^33x+1=0\) has a root in the interval \((0,1)\).
Problem Statement 

Use the intermediate value theorem to solve this problem. Give your answer in exact form.
Show that \(x^33x+1=0\) has a root in the interval \((0,1)\).
Solution 

video by PatrickJMT 

close solution

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Show that \(f(x)=x^4+x3\) has a root in the interval \((1,2)\).
Problem Statement 

Use the intermediate value theorem to solve this problem. Give your answer in exact form.
Show that \(f(x)=x^4+x3\) has a root in the interval \((1,2)\).
Solution 

video by Krista King Math 

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Show that \(x^2=\sqrt{x+1}\) has a root in the interval \((1,2)\).
Problem Statement 

Use the intermediate value theorem to solve this problem. Give your answer in exact form.
Show that \(x^2=\sqrt{x+1}\) has a root in the interval \((1,2)\).
Solution 

video by PatrickJMT 

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Show that \(e^x=2\cos(x)\) has at least one positive root.
Problem Statement 

Use the intermediate value theorem to solve this problem. Give your answer in exact form.
Show that \(e^x=2\cos(x)\) has at least one positive root.
Solution 

He chose \(x=0\) and \(x=\pi/2\) to show that the function has a positive value and negative value. You can choose other values as long as one results in a positive value and the other is negative.
video by Dr Chris Tisdell 

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Prove that \(\cos(x)=x^3\) has at least one real root.
Problem Statement 

Use the intermediate value theorem to solve this problem. Give your answer in exact form.
Prove that \(\cos(x)=x^3\) has at least one real root.
Solution 

video by Krista King Math 

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external links you may find helpful 

The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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Practice Instructions
Use the intermediate value theorem to solve these problems. Give your answers in exact form.