While the idea of continuity may seem somewhat basic, when a function is continuous over a closed interval like \begin{align*}x \in [1,4]\end{align*}
What can you conclude using the Intermediate Value Theorem and the Extreme Value Theorem about a function that is continuous over the closed interval \begin{align*}x \in [1,4]\end{align*}
Watch This
http://www.youtube.com/watch?v=6AFT1wnId9U Intermediate Value Theorem
Guidance
The Intermediate Value Theorem states that if a function is continuous on a closed interval and \begin{align*}u\end{align*}
Simply stated, if a function is continuous between a low point and a high point, then it must be valued at each intermediate height in between the low and high points.
The Extreme Value Theorem states that in every interval \begin{align*}[a,b]\end{align*}
Example A
Show that the converse of the Intermediate Value Theorem is false.
Solution: The converse of the Intermediate Value Theorem is: If there exists a value \begin{align*}c\in[a,b]\end{align*}
In order to show the statement is false, all you need is one counterexample where every intermediate value is hit and the function is discontinuous.
This function is discontinuous on the interval \begin{align*}[0,10]\end{align*}
Example B
Show that the converse of the Extreme Value Theorem is false.
Solution: The converse of the Extreme Value Theorem is: If there is at least one maximum and one minimum in the closed interval \begin{align*}[a,b]\end{align*}
In order to show the statement is false, all you need is one counterexample. The goal is to find a function on a closed interval \begin{align*}[a,b]\end{align*}
On the interval \begin{align*}[0,10]\end{align*}
Example C
Use the Intermediate Value Theorem to show that the function \begin{align*}f(x)=(x+1)^34\end{align*}
Solution: First note that the function is a cubic and is therefore continuous everywhere.

\begin{align*}f(0)=(0+1)^34=1^34=3\end{align*}
f(0)=(0+1)3−4=13−4=−3 
\begin{align*}f(3)=(3+1)^34=4^34=60\end{align*}
f(3)=(3+1)3−4=43−4=60
By the Intermediate Value Theorem, there must exist a \begin{align*}c \in [0,3]\end{align*}
Concept Problem Revisited
If a function is continuous on the interval \begin{align*}x \in [1,4]\end{align*}
Vocabulary
The converse of an if then statement is a new statement with the hypothesis of the original statement switched with the conclusion of the original statement. In other words, the converse is when the if part of the statement and the then part of the statement are swapped. In general, the converse of a statement is not true.
A counterexample to an if then statement is when the hypothesis (the if part of the sentence) is true, but the conclusion (the then part of the statement) is not true.
Guided Practice
1. Use the Intermediate Value Theorem to show that the following equation has at least one real solution.
\begin{align*}x^8=2^x\end{align*}
2. Show that there is at least one solution to the following equation.
\begin{align*}\sin x=x+2\end{align*}
3. When are you not allowed to use the Intermediate Value Theorem?
Answers:
1. First rewrite the equation: \begin{align*}x^82^x=0\end{align*}
Then describe it as a continuous function: \begin{align*}f(x)=x^82^x\end{align*}
This function is continuous because it is the difference of two continuous functions.

\begin{align*}f(0)=0^82^0=01=1 \end{align*}
f(0)=08−20=0−1=−1 
\begin{align*}f(2)=2^82^2=2564=252\end{align*}
f(2)=28−22=256−4=252
By the Intermediate Value Theorem, there must exist a \begin{align*}c\end{align*}
2. Write the equation as a continuous function: \begin{align*}f(x)=\sin xx2\end{align*}
The function is continuous because it is the sum and difference of continuous functions.
 \begin{align*}f(0)=\sin 002=2\end{align*}
 \begin{align*}f(\pi)=\sin(\pi)+\pi 2=0+\pi 2 > 0\end{align*}
By the Intermediate Value Theorem, there must exist a \begin{align*}c\end{align*} such that \begin{align*}f(c)=0\end{align*} because \begin{align*}2 < 0 < \pi 2\end{align*}. The number \begin{align*}c\end{align*} is one solution to the initial equation.
3. The Intermediate Value Theorem should not be applied when the function is not continuous over the interval.
Practice
Use the Intermediate Value Theorem to show that each equation has at least one real solution.
1. \begin{align*}\cos x=x\end{align*}
2. \begin{align*}\ln (x)=e^{x}+1\end{align*}
3. \begin{align*}2x^35x^2=10x5\end{align*}
4. \begin{align*}x^3+1=x\end{align*}
5. \begin{align*}x^2=\cos x\end{align*}
6. \begin{align*}x^5=2x^3+2\end{align*}
7. \begin{align*}3x^2+4x11=0\end{align*}
8. \begin{align*}5x^4=6x^2+1\end{align*}
9. \begin{align*}7x^318x^24x+1=0\end{align*}
10. Show that \begin{align*}f(x)=\frac{2x3}{2x5}\end{align*} has a real root on the interval \begin{align*}[1,2]\end{align*}.
11. Show that \begin{align*}f(x)=\frac{3x+1}{2x+4}\end{align*} has a real root on the interval \begin{align*}[1,0]\end{align*}.
12. True or false: A function has a maximum and a minimum in the closed interval \begin{align*}[a,b]\end{align*}; therefore, the function is continuous.
13. True or false: A function is continuous over the interval \begin{align*}[a,b]\end{align*}; therefore, the function has a maximum and a minimum in the closed interval.
14. True or false: If a function is continuous over the interval \begin{align*}[a,b]\end{align*}, then it is possible for the function to have more than one relative maximum in the interval \begin{align*}[a,b]\end{align*}.
15. What do the Intermediate Value and Extreme Value Theorems have to do with continuity?