Difference between revisions of "009A Sample Final 1, Problem 6"

Revision as of 10:29, 15 February 2016

Consider the following function:

f(x)=3x-2\sin x+7

a) Use the Intermediate Value Theorem to show that $f(x)$ has at least one zero.

b) Use the Mean Value Theorem to show that $f(x)$ has at most one zero.

Foundations:

Solution:

(a)

Step 2:
Since $f(-5)<0$ and $f(0)>0$ , there exists $x$ with $-5<x<0$ such that
$f(x)=0$ by the Intermediate Value Theorem. Hence, $f(x)$ has at least one zero.

(b)

Step 2:

Step 3:

Final Answer:
(a) Since $f(-5)<0$ and $f(0)>0$ , there exists $x$ with $-5<x<0$ such that
$f(x)=0$ by the Intermediate Value Theorem. Hence, $f(x)$ has at least one zero.
(b)

@@ Line 20: / Line 20: @@
 !Step 1: &nbsp;
 |-
-|
+|First note that <math>f(0)=7</math>.
+|-
+|Also, <math>f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5)</math>.
 |-
-|
+|Since <math>-1\leq \sin(x) \leq 1</math>,
 |-
-|
+|<math>-2\leq -2\sin(x) \leq 2</math>.
 |-
-|
+|Thus, <math>-10\leq f(-5) \leq -6</math> and hence <math>f(-5)<0</math>.
 |}
@@ Line 32: / Line 34: @@
 !Step 2: &nbsp;
 |-
-|
+|Since <math>f(-5)<0</math> and <math>f(0)>0</math>, there exists <math>x</math> with <math>-5<x<0</math> such that
 |-
-|
+|<math>f(x)=0</math> by the Intermediate Value Theorem. Hence, <math>f(x)</math> has at least one zero.
-|-
-|
 |}
@@ Line 68: / Line 68: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|'''(a)''' Since <math>f(-5)<0</math> and <math>f(0)>0</math>, there exists <math>x</math> with <math>-5<x<0</math> such that
+|-
+|<math>f(x)=0</math> by the Intermediate Value Theorem. Hence, <math>f(x)</math> has at least one zero.
 |-
 |'''(b)'''
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]