Difference between revisions of "009A Sample Midterm 2, Problem 2"

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!Foundations:    
 
!Foundations:    
 
|-
 
|-
|What is a zero of the function <math style="vertical-align: -5px">f(x)?</math>
+
|What is a zero of the function &nbsp;<math style="vertical-align: -5px">f(x)?</math>
 
|-
 
|-
|&nbsp; &nbsp; &nbsp; &nbsp; A zero is a value <math style="vertical-align: -1px">c</math> such that <math style="vertical-align: -5px">f(c)=0.</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp; A zero is a value &nbsp;<math style="vertical-align: -1px">c</math>&nbsp; such that &nbsp;<math style="vertical-align: -5px">f(c)=0.</math>
 
|}
 
|}
  
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|'''Intermediate Value Theorem'''  
 
|'''Intermediate Value Theorem'''  
 
|-
 
|-
|&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -5px">f(x)</math>&thinsp; is continuous on a closed interval <math style="vertical-align: -5px">[a,b]</math>  
+
|&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous on a closed interval &nbsp;<math style="vertical-align: -5px">[a,b]</math>  
 
|-
 
|-
|&nbsp; &nbsp; &nbsp; &nbsp; and <math style="vertical-align: 0px">c</math> is any number between <math style="vertical-align: -5px">f(a)</math>&thinsp; and <math style="vertical-align: -5px">f(b),</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp;<math style="vertical-align: 0px">c</math>&nbsp; is any number between &nbsp;<math style="vertical-align: -5px">f(a)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(b),</math>
 
|-
 
|-
 
|
 
|
&nbsp; &nbsp; &nbsp; &nbsp; then there is at least one number <math style="vertical-align: 0px">x</math> in the closed interval such that <math style="vertical-align: -5px">f(x)=c.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; then there is at least one number &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; in the closed interval such that &nbsp;<math style="vertical-align: -5px">f(x)=c.</math>
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|First, <math style="vertical-align: -5px">f(x)</math> is continuous on the interval <math style="vertical-align: -5px">[0,1]</math> since <math style="vertical-align: -5px">f(x)</math> is continuous everywhere.
+
|First, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous on the interval &nbsp;<math style="vertical-align: -5px">[0,1]</math>&nbsp; since &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous everywhere.
 
|-
 
|-
 
|Also,
 
|Also,
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|Since <math style="vertical-align: -1px">0</math> is between <math style="vertical-align: -5px">f(0)=2</math>&nbsp; and <math style="vertical-align: -5px">f(1)=-3,</math>
+
|Since &nbsp;<math style="vertical-align: -1px">0</math>&nbsp; is between &nbsp;<math style="vertical-align: -5px">f(0)=2</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(1)=-3,</math>
 
|-
 
|-
|the Intermediate Value Theorem tells us that there is at least one number <math style="vertical-align: -1px">x</math>
+
|the Intermediate Value Theorem tells us that there is at least one number &nbsp;<math style="vertical-align: -1px">x</math>
 
|-
 
|-
|such that <math style="vertical-align: -5px">f(x)=0.</math>
+
|such that &nbsp;<math style="vertical-align: -5px">f(x)=0.</math>
 
|-
 
|-
|This means that <math style="vertical-align: -5px">f(x)</math> has a zero in the interval <math style="vertical-align: -5px">[0,1].</math>
+
|This means that &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has a zero in the interval &nbsp;<math style="vertical-align: -5px">[0,1].</math>
 
|}
 
|}
  

Revision as of 17:02, 26 February 2017

The function    is a polynomial and therefore continuous everywhere.

(a) State the Intermediate Value Theorem.

(b) Use the Intermediate Value Theorem to show that    has a zero in the interval  


Foundations:  
What is a zero of the function  
        A zero is a value    such that  


Solution:

(a)  
Intermediate Value Theorem
        If    is continuous on a closed interval  
        and    is any number between    and  

        then there is at least one number    in the closed interval such that  

(b)

Step 1:  
First,    is continuous on the interval    since    is continuous everywhere.
Also,

       

and

        .

Step 2:  
Since    is between    and  
the Intermediate Value Theorem tells us that there is at least one number  
such that  
This means that    has a zero in the interval  


Final Answer:  
    (a)     See solution above.
    (b)     See solution above.

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