14.7. Intermediate and Extreme Value Theorems http://www.ck12.org
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[a,b]then the function is continuous on[a,b].
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[a,b]that has at least one maximum and one minimum and is also discontinuous.
On the interval[ 0 , 10 ], the function attains a maximum at( 5 , 5 )and a minimum at( 0 , 0 )but is still discontinuous.
Example C
Use the Intermediate Value Theorem to show that the functionf(x) = (x+ 1 )^3 −4 has a zero on the interval[ 0 , 3 ].
Solution:First note that the function is a cubic and is therefore continuous everywhere.
- f( 0 ) = ( 0 + 1 )^3 − 4 = 13 − 4 =− 3
- f( 3 ) = ( 3 + 1 )^3 − 4 = 43 − 4 = 60
By the Intermediate Value Theorem, there must exist ac∈[ 0 , 3 ]such thatf(c) =0 since 0 is between -3 and 60.
Concept Problem Revisited
If a function is continuous on the intervalx∈[ 1 , 4 ], then you can conclude by the Intermediate Value Theorem that
there exists ac∈[ 1 , 4 ]such thatf(c) =ufor everyubetweenf( 1 )andf( 4 ). You can also conclude that on this
interval the function has both a maximum and a minimum value.
Vocabulary
Theconverseof 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.
Acounterexampleto 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
- Use the Intermediate Value Theorem to show that the following equation has at least one real solution.