009A Sample Final 1, Problem 6 - Revision history

Kayla Murray at 22:53, 2 December 2017

2017-12-02T22:53:11Z

Kayla Murray at 20:19, 18 March 2017

2017-03-18T20:19:30Z

Kayla Murray at 16:27, 27 February 2017

2017-02-27T16:27:51Z

Kayla Murray at 00:48, 26 February 2017

2017-02-26T00:48:56Z

Kayla Murray at 01:48, 19 February 2017

2017-02-19T01:48:19Z

Kayla Murray at 01:43, 19 February 2017

2017-02-19T01:43:49Z

Grad at 19:15, 4 March 2016

2016-03-04T19:15:25Z

Grad: /* 2 */

2016-03-04T19:15:07Z

Grad: /* 3 */

2016-03-04T19:12:51Z

Grad: /* 1 */

2016-03-04T19:11:15Z

@@ Line 7: / Line 7: @@
 <span class="exam">(b) Use the Mean Value Theorem to show that &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has at most one zero.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Final 1, Problem 6 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009A Sample Final 1, Problem 6 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 64: / Line 64: @@
 !Step 1: &nbsp;
 |-
-|Suppose that &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has more than one zero. So, there exist &nbsp;<math style="vertical-align: -4px">a,b</math>&nbsp; such that &nbsp;<math style="vertical-align: -5px">f(a)=f(b)=0.</math>
+|Suppose that &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has more than one zero. So, there exist &nbsp;<math style="vertical-align: -4px">a,b</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">a<b</math>&nbsp; such that &nbsp;<math style="vertical-align: -5px">f(a)=f(b)=0.</math>
 |-
 |Then, by the Mean Value Theorem, there exists &nbsp;<math style="vertical-align: 0px">c</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">a<c<b</math>&nbsp; such that &nbsp;<math style="vertical-align: -5px">f'(c)=0.</math>
@@ Line 72: / Line 72: @@
 !Step 2: &nbsp;
 |-
-|We have &nbsp;<math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math>&nbsp; Since &nbsp;<math style="vertical-align: -5px">-1\leq \cos(x)\leq 1,</math>
+|We have &nbsp;<math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math>&nbsp;
 |-
-|<math style="vertical-align: -5px">-2 \leq -2\cos(x)\leq 2.</math>&nbsp; So, &nbsp;<math style="vertical-align: -5px">1\leq f'(x) \leq 5,</math>
+|Since &nbsp;<math style="vertical-align: -5px">-1\leq \cos(x)\leq 1,</math>
 |-
 |which contradicts &nbsp;<math style="vertical-align: -5px">f'(c)=0.</math>&nbsp; Thus, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has at most one zero.
@@ Line 83: / Line 87: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; Since &nbsp;<math style="vertical-align: -5px">f(-5)<0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(0)>0,</math>&nbsp; there exists &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">-5<x<0</math>&nbsp; such that
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; See solution above.
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; See Step 1 and Step 2 above.
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; See solution above.
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 3: / Line 3: @@
 ::<math>f(x)=3x-2\sin x+7</math>
-<span class="exam">(a) Use the Intermediate Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>&thinsp; has at least one zero.
+<span class="exam">(a) Use the Intermediate Value Theorem to show that &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has at least one zero.
-<span class="exam">(b) Use the Mean Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>&thinsp; has at most one zero.
+<span class="exam">(b) Use the Mean Value Theorem to show that &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has at most one zero.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 12: / Line 12: @@
 |'''1.''' '''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> and <math style="vertical-align: 0px">c</math> is any number
+|&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; and &nbsp;<math style="vertical-align: 0px">c</math>&nbsp; is any number
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;between <math style="vertical-align: -5px">f(a)</math>&thinsp; and <math style="vertical-align: -5px">f(b)</math>, 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;between &nbsp;<math style="vertical-align: -5px">f(a)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(b),</math>&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>
 |-
 |'''2.'''  '''Mean Value Theorem'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Suppose <math style="vertical-align: -5px">f(x)</math>&thinsp; is a function that satisfies the following:
+|&nbsp; &nbsp; &nbsp; &nbsp; Suppose &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is a function that satisfies the following:
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&thinsp; is continuous on the closed interval &thinsp;<math style="vertical-align: -5px">[a,b].</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous on the closed interval &nbsp;<math style="vertical-align: -5px">[a,b].</math>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&thinsp; is differentiable on the open interval <math style="vertical-align: -5px">(a,b).</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is differentiable on the open interval &nbsp;<math style="vertical-align: -5px">(a,b).</math>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;Then, there is a number <math style="vertical-align: 0px">c</math> such that &thinsp;<math style="vertical-align: 0px">a<c<b</math>&thinsp; and <math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;Then, there is a number &nbsp;<math style="vertical-align: 0px">c</math>&nbsp; such that &nbsp;<math style="vertical-align: 0px">a<c<b</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
 |}
@@ Line 41: / Line 41: @@
 |First note that &nbsp; <math style="vertical-align: -5px">f(0)=7.</math>
 |-
-|Also,&nbsp; <math style="vertical-align: -5px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).</math>
+|Also, &nbsp;<math style="vertical-align: -5px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).</math>
 |-
-|Since&nbsp; <math style="vertical-align: -5px">-1\leq \sin(x) \leq 1,</math>
+|Since &nbsp;<math style="vertical-align: -5px">-1\leq \sin(x) \leq 1,</math>
 |-
+|
 &nbsp; &nbsp; &nbsp; &nbsp;<math>-2\leq -2\sin(x) \leq 2.</math>
 |-
-|Thus,&nbsp; <math style="vertical-align: -5px">-10\leq f(-5) \leq -6</math>&nbsp; and hence &nbsp;<math style="vertical-align: -5px">f(-5)<0.</math>
+|Thus, &nbsp;<math style="vertical-align: -5px">-10\leq f(-5) \leq -6</math>&nbsp; and hence &nbsp;<math style="vertical-align: -5px">f(-5)<0.</math>
 |}
@@ Line 54: / Line 54: @@
 !Step 2: &nbsp;
 |-
-|Since <math style="vertical-align: -5px">f(-5)<0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(0)>0,</math>&nbsp; there exists <math style="vertical-align: 0px">x</math> with &nbsp;<math style="vertical-align: 0px">-5<x<0</math>&nbsp; such that
+|Since &nbsp;<math style="vertical-align: -5px">f(-5)<0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(0)>0,</math>&nbsp; there exists &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">-5<x<0</math>&nbsp; such that
 |-
-|<math style="vertical-align: -5px">f(x)=0</math>&nbsp; by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>&nbsp; has at least one zero.
+|<math style="vertical-align: -5px">f(x)=0</math>&nbsp; by the Intermediate Value Theorem. Hence, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has at least one zero.
 |}
@@ Line 64: / Line 64: @@
 !Step 1: &nbsp;
 |-
-|Suppose that <math style="vertical-align: -5px">f(x)</math> has more than one zero. So, there exist <math style="vertical-align: -4px">a,b</math> such that &nbsp;<math style="vertical-align: -5px">f(a)=f(b)=0.</math>
+|Suppose that &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has more than one zero. So, there exist &nbsp;<math style="vertical-align: -4px">a,b</math>&nbsp; such that &nbsp;<math style="vertical-align: -5px">f(a)=f(b)=0.</math>
 |-
-|Then, by the Mean Value Theorem, there exists <math style="vertical-align: 0px">c</math> with &nbsp;<math style="vertical-align: 0px">a<c<b</math> such that &nbsp;<math style="vertical-align: -5px">f'(c)=0.</math>
+|Then, by the Mean Value Theorem, there exists &nbsp;<math style="vertical-align: 0px">c</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">a<c<b</math>&nbsp; such that &nbsp;<math style="vertical-align: -5px">f'(c)=0.</math>
 |}
@@ Line 72: / Line 72: @@
 !Step 2: &nbsp;
 |-
-|We have <math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math>&nbsp; Since &nbsp;<math style="vertical-align: -5px">-1\leq \cos(x)\leq 1,</math>
+|We have &nbsp;<math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math>&nbsp; Since &nbsp;<math style="vertical-align: -5px">-1\leq \cos(x)\leq 1,</math>
 |-
-|<math style="vertical-align: -5px">-2 \leq -2\cos(x)\leq 2.</math>&nbsp; So, <math style="vertical-align: -5px">1\leq f'(x) \leq 5,</math>
+|<math style="vertical-align: -5px">-2 \leq -2\cos(x)\leq 2.</math>&nbsp; So, &nbsp;<math style="vertical-align: -5px">1\leq f'(x) \leq 5,</math>
 |-
-|which contradicts <math style="vertical-align: -5px">f'(c)=0.</math> Thus, <math style="vertical-align: -5px">f(x)</math>&nbsp; has at most one zero.
+|which contradicts &nbsp;<math style="vertical-align: -5px">f'(c)=0.</math>&nbsp; Thus, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has at most one zero.
 |}
@@ Line 83: / Line 83: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; Since <math style="vertical-align: -5px">f(-5)<0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(0)>0,</math>&nbsp; there exists <math style="vertical-align: 0px">x</math> with &nbsp;<math style="vertical-align: 0px">-5<x<0</math>&nbsp; such that
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; Since &nbsp;<math style="vertical-align: -5px">f(-5)<0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(0)>0,</math>&nbsp; there exists &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">-5<x<0</math>&nbsp; such that
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">f(x)=0</math>&nbsp; by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>&nbsp; has at least one zero.
+|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">f(x)=0</math>&nbsp; by the Intermediate Value Theorem. Hence, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has at least one zero.
 |-
 |&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; See Step 1 and Step 2 above.
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 10: / Line 10: @@
 !Foundations: &nbsp;
 |-
-|Recall:
+|'''1.''' '''Intermediate Value Theorem'''
 |-
-|'''1. Intermediate Value Theorem:''' If <math style="vertical-align: -5px">f(x)</math>&thinsp; is continuous on a closed interval <math style="vertical-align: -5px">[a,b]</math> and <math style="vertical-align: 0px">c</math> is any number
+|&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> 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>, 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;between <math style="vertical-align: -5px">f(a)</math>&thinsp; and <math style="vertical-align: -5px">f(b)</math>, 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>
 |-
-|'''2. Mean Value Theorem:''' Suppose <math style="vertical-align: -5px">f(x)</math>&thinsp; is a function that satisfies the following:
+|'''2.'''  '''Mean Value Theorem'''
 |-
+|
-::<math style="vertical-align: -5px">f(x)</math>&thinsp; is continuous on the closed interval &thinsp;<math style="vertical-align: -5px">[a,b].</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&thinsp; is continuous on the closed interval &thinsp;<math style="vertical-align: -5px">[a,b].</math>
 |-
+|
-::<math style="vertical-align: -5px">f(x)</math>&thinsp; is differentiable on the open interval <math style="vertical-align: -5px">(a,b).</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&thinsp; is differentiable on the open interval <math style="vertical-align: -5px">(a,b).</math>
 |-
+|
-::Then, there is a number <math style="vertical-align: 0px">c</math> such that &thinsp;<math style="vertical-align: 0px">a<c<b</math>&thinsp; and <math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;Then, there is a number <math style="vertical-align: 0px">c</math> such that &thinsp;<math style="vertical-align: 0px">a<c<b</math>&thinsp; and <math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
 |}
@@ Line 37: / Line 39: @@
 !Step 1: &nbsp;
 |-
-|First note that&thinsp; <math style="vertical-align: -5px">f(0)=7.</math>
+|First note that &nbsp; <math style="vertical-align: -5px">f(0)=7.</math>
 |-
-|Also,&thinsp; <math style="vertical-align: -5px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).</math>
+|Also,&nbsp; <math style="vertical-align: -5px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).</math>
 |-
-|Since&thinsp; <math style="vertical-align: -5px">-1\leq \sin(x) \leq 1,</math>
+|Since&nbsp; <math style="vertical-align: -5px">-1\leq \sin(x) \leq 1,</math>
 |-
+|
-::<math>-2\leq -2\sin(x) \leq 2.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>-2\leq -2\sin(x) \leq 2.</math>
 |-
-|Thus,&thinsp; <math style="vertical-align: -5px">-10\leq f(-5) \leq -6</math>&thinsp; and hence &thinsp;<math style="vertical-align: -5px">f(-5)<0.</math>
+|Thus,&nbsp; <math style="vertical-align: -5px">-10\leq f(-5) \leq -6</math>&nbsp; and hence &nbsp;<math style="vertical-align: -5px">f(-5)<0.</math>
 |}
@@ Line 52: / Line 54: @@
 !Step 2: &nbsp;
 |-
-|Since <math style="vertical-align: -5px">f(-5)<0</math>&thinsp; and &thinsp;<math style="vertical-align: -5px">f(0)>0,</math>&thinsp; there exists <math style="vertical-align: 0px">x</math> with &thinsp;<math style="vertical-align: 0px">-5<x<0</math>&thinsp; such that
+|Since <math style="vertical-align: -5px">f(-5)<0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(0)>0,</math>&nbsp; there exists <math style="vertical-align: 0px">x</math> with &nbsp;<math style="vertical-align: 0px">-5<x<0</math>&nbsp; such that
 |-
-|<math style="vertical-align: -5px">f(x)=0</math>&thinsp; by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>&thinsp; has at least one zero.
+|<math style="vertical-align: -5px">f(x)=0</math>&nbsp; by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>&nbsp; has at least one zero.
 |}
@@ Line 62: / Line 64: @@
 !Step 1: &nbsp;
 |-
-|Suppose that <math style="vertical-align: -5px">f(x)</math> has more than one zero. So, there exist <math style="vertical-align: -4px">a,b</math> such that &thinsp;<math style="vertical-align: -5px">f(a)=f(b)=0.</math>
+|Suppose that <math style="vertical-align: -5px">f(x)</math> has more than one zero. So, there exist <math style="vertical-align: -4px">a,b</math> such that &nbsp;<math style="vertical-align: -5px">f(a)=f(b)=0.</math>
 |-
-|Then, by the Mean Value Theorem, there exists <math style="vertical-align: 0px">c</math> with &thinsp;<math style="vertical-align: 0px">a<c<b</math> such that &thinsp;<math style="vertical-align: -5px">f'(c)=0.</math>
+|Then, by the Mean Value Theorem, there exists <math style="vertical-align: 0px">c</math> with &nbsp;<math style="vertical-align: 0px">a<c<b</math> such that &nbsp;<math style="vertical-align: -5px">f'(c)=0.</math>
 |}
@@ Line 70: / Line 72: @@
 !Step 2: &nbsp;
 |-
-|We have <math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math>&thinsp; Since &thinsp;<math style="vertical-align: -5px">-1\leq \cos(x)\leq 1,</math>
+|We have <math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math>&nbsp; Since &nbsp;<math style="vertical-align: -5px">-1\leq \cos(x)\leq 1,</math>
 |-
-|<math style="vertical-align: -5px">-2 \leq -2\cos(x)\leq 2.</math>&thinsp; So, <math style="vertical-align: -5px">1\leq f'(x) \leq 5,</math>
+|<math style="vertical-align: -5px">-2 \leq -2\cos(x)\leq 2.</math>&nbsp; So, <math style="vertical-align: -5px">1\leq f'(x) \leq 5,</math>
 |-
-|which contradicts <math style="vertical-align: -5px">f'(c)=0.</math> Thus, <math style="vertical-align: -5px">f(x)</math>&thinsp; has at most one zero.
+|which contradicts <math style="vertical-align: -5px">f'(c)=0.</math> Thus, <math style="vertical-align: -5px">f(x)</math>&nbsp; has at most one zero.
 |}
@@ Line 81: / Line 83: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' Since <math style="vertical-align: -5px">f(-5)<0</math>&thinsp; and &thinsp;<math style="vertical-align: -5px">f(0)>0,</math>&thinsp; there exists <math style="vertical-align: 0px">x</math> with &thinsp;<math style="vertical-align: 0px">-5<x<0</math>&thinsp; such that
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; Since <math style="vertical-align: -5px">f(-5)<0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(0)>0,</math>&nbsp; there exists <math style="vertical-align: 0px">x</math> with &nbsp;<math style="vertical-align: 0px">-5<x<0</math>&nbsp; such that
 |-
-|<math style="vertical-align: -5px">f(x)=0</math>&thinsp; by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>&thinsp; has at least one zero.
+|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">f(x)=0</math>&nbsp; by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>&nbsp; has at least one zero.
 |-
-|'''(b)''' See '''Step 1''' and '''Step 2''' above.
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; See Step 1 and Step 2 above.
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 28: / Line 28: @@
 ::Then, there is a number <math style="vertical-align: 0px">c</math> such that &thinsp;<math style="vertical-align: 0px">a<c<b</math>&thinsp; and <math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
 |}
 '''Solution:'''
@@ Line 75: / Line 76: @@
 |which contradicts <math style="vertical-align: -5px">f'(c)=0.</math> Thus, <math style="vertical-align: -5px">f(x)</math>&thinsp; has at most one zero.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 1: / Line 1: @@
 <span class="exam"> Consider the following function:
-::::::<math>f(x)=3x-2\sin x+7</math>
+::<math>f(x)=3x-2\sin x+7</math>
-<span class="exam">a) Use the Intermediate Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>&thinsp; has at least one zero.
+<span class="exam">(a) Use the Intermediate Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>&thinsp; has at least one zero.
-<span class="exam">b) Use the Mean Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>&thinsp; has at most one zero.
+<span class="exam">(b) Use the Mean Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>&thinsp; has at most one zero.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"