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

Revision as of 13:13, 17 February 2017

The function $f(x)=3x^{7}-8x+2$ is a polynomial and therefore continuous everywhere.

a) State the Intermediate Value Theorem.

b) Use the Intermediate Value Theorem to show that

f(x)

has a zero in the interval

[0,1].

Foundations:
?

Solution:

(a)

Step 1:
Intermediate Value Theorem
If $f(x)$ is continuous on a closed interval $[a,b]$
and $c$ is any number between $f(a)$ and $f(b)$ ,
then there is at least one number $x$ in the closed interval such that $f(x)=c.$

(b)

Step 1:
First, $f(x)$ is continuous on the interval $[0,1]$ since $f(x)$ is continuous everywhere.
Also,
$f(0)=2$
and $f(1)=3-8+2=-3.$ .

Step 2:
Since $0$ is between $f(0)=2$ and $f(1)=-3,$
the Intermediate Value Theorem tells us that there is at least one number $x$
such that $f(x)=0.$
This means that $f(x)$ has a zero in the interval $[0,1].$

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

@@ Line 31: / Line 31: @@
 !Step 1: &nbsp;
 |-
-|
+|First, <math>f(x)</math> is continuous on the interval <math>[0,1]</math> since <math>f(x)</math> is continuous everywhere.
 |-
-|
+|Also,
 |-
 |
+&nbsp; &nbsp; &nbsp; &nbsp; <math>f(0)=2</math>
 |-
-|
+|and
+&nbsp; &nbsp; &nbsp; &nbsp; <math>f(1)=3-8+2=-3.</math>.
 |}
@@ Line 43: / Line 45: @@
 !Step 2: &nbsp;
 |-
-|
+|Since <math>0</math> is between <math>f(0)=2</math> and <math>f(1)=-3,</math>
 |-
-|
+|the Intermediate Value Theorem tells us that there is at least one number <math>x</math>
 |-
-|
+|such that <math>f(x)=0.</math>
 |-
-|
+|This means that <math>f(x)</math> has a zero in the interval <math>[0,1].</math>
 |}
@@ Line 56: / Line 58: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; See solution above.
 |-
-|'''(b)'''
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; See solution above.
 |}
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]