Math 22 Continuity - Revision history

Tphan046 at 13:51, 23 July 2020

2020-07-23T13:51:52Z

Tphan046: /* Notes */

2020-07-19T14:51:19Z

Notes

Tphan046: /* Continuity of piece-wise functions */

2020-07-16T16:22:55Z

Continuity of piece-wise functions

Tphan046: /* Continuity of piece-wise functions */

2020-07-16T16:22:28Z

Continuity of piece-wise functions

Tphan046: /* Continuity of piece-wise functions */

2020-07-16T16:20:35Z

Continuity of piece-wise functions

Tphan046: /* Continuity of piece-wise functions */

2020-07-16T16:20:09Z

Continuity of piece-wise functions

Tphan046: /* Notes */

2020-07-16T16:12:02Z

Notes

Tphan046: /* Definition of Continuity */

2020-07-16T16:11:42Z

Definition of Continuity

Tphan046: /* Continuity */

2020-07-16T16:11:13Z

Continuity

Tphan046: /* Continuity of piece-wise functions */

2020-07-16T16:10:59Z

Continuity of piece-wise functions

@@ Line 38: / Line 38: @@
 |Hence, <math>f(x)</math> is continuous on <math>[-1,5]</math>
 |}
 ==Notes==

@@ Line 45: / Line 45: @@
 ''Rational functions'' is continuous at every number in its domain. (ex: <math>f(x)=\frac {x+2}{x^2-1}</math> is continuous on <math>(-\infty,-1)\cup (-1,1)\cup (1,\infty)</math> since the denominator cannot equal to zero)
+[[Math_22| '''Return to Topics Page''']]
 '''This page were made by [[Contributors|Tri Phan]]'''

@@ Line 19: / Line 19: @@
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 !Solution: &nbsp;
 |On the interval <math>[-1,3)</math>, <math>f(x)=x+2</math> and it is a polynomial function so it is continuous on <math>[-1,3)</math>
 |-

@@ Line 17: / Line 17: @@
 \end{cases}</math>
-On the interval <math>[-1,3)</math>, <math>f(x)=x+2</math> and it is a polynomial function so it is continuous on <math>[-1,3)</math>
+{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
+!Solution: &nbsp;
-On the interval <math>[3,5]</math>, <math>f(x)=14-x^2</math> and it is a polynomial function so it is continuous on <math>[3,5]</math>
+|On the interval <math>[-1,3)</math>, <math>f(x)=x+2</math> and it is a polynomial function so it is continuous on <math>[-1,3)</math>
+|-
-Finally we need to check if <math>f(x)</math> is continuous at <math>x=3</math>.
+|On the interval <math>[3,5]</math>, <math>f(x)=14-x^2</math> and it is a polynomial function so it is continuous on <math>[3,5]</math>
+|-
-So, consider <math>\lim_{x\to 3^-} f(x)= \lim_{x\to 3^-} x+2= 3+2=5</math>
+|Finally we need to check if <math>f(x)</math> is continuous at <math>x=3</math>.
+|-
-Then, <math>\lim_{x\to 3^+} f(x)= \lim_{x\to 3^+} 14-x^2=14-(3)^2=5</math>.
+|So, consider <math>\lim_{x\to 3^-} f(x)= \lim_{x\to 3^-} x+2= 3+2=5</math>
+|-
-Since <math>\lim_{x\to 3^-} f(x)= 5 = \lim_{x\to 3^+} f(x)</math>, \lim_{x\to 3} f(x) exists.
+|Then, <math>\lim_{x\to 3^+} f(x)= \lim_{x\to 3^+} 14-x^2=14-(3)^2=5</math>.
+|-
-Also notice <math>f(3)=14-(3)^2=5=\lim_{x\to 3} f(x)</math>
+|Since <math>\lim_{x\to 3^-} f(x)= 5 = \lim_{x\to 3^+} f(x)</math>, \lim_{x\to 3} f(x) exists.
+|-
-So by definition of continuity, <math>f(x)</math> is continuous at <math>x=3</math>.
+|Also notice <math>f(3)=14-(3)^2=5=\lim_{x\to 3} f(x)</math>
+|-
-Hence, <math>f(x)</math> is continuous on <math>[-1,5]</math>
+|So by definition of continuity, <math>f(x)</math> is continuous at <math>x=3</math>.
 ==Notes==

@@ Line 27: / Line 27: @@
 Then, <math>\lim_{x\to 3^+} f(x)= \lim_{x\to 3^+} 14-x^2=14-(3)^2=5</math>.
-Since <math>\lim_{x\to 3^-} f(x)= 5 = \lim_{x\to 3^+} f(x)</math>, \lim_{x\to 3^} f(x) exists.
+Since <math>\lim_{x\to 3^-} f(x)= 5 = \lim_{x\to 3^+} f(x)</math>, \lim_{x\to 3} f(x) exists.
-Also notice <math>f(3)=14-(3)^2=5=\lim_{x\to 3^} f(x)</math>
+Also notice <math>f(3)=14-(3)^2=5=\lim_{x\to 3} f(x)</math>
 So by definition of continuity, <math>f(x)</math> is continuous at <math>x=3</math>.

@@ Line 21: / Line 21: @@
 ''Polynomial function'' is continuous on the entire real number line (ex: <math>f(x)=2x^2-1</math> is continuous on <math>(-\infty,\infty)</math>)
-''Rational Functions'' is continuous at every number in its domain. (ex: <math>f(x)=\frac {x+2}{x^2-1}</math> is continuous on <math>(-\infty,-1)\cup (-1,1)\cup (1,\infty)</math> since the denominator cannot equal to zero)
+''Rational functions'' is continuous at every number in its domain. (ex: <math>f(x)=\frac {x+2}{x^2-1}</math> is continuous on <math>(-\infty,-1)\cup (-1,1)\cup (1,\infty)</math> since the denominator cannot equal to zero)
 '''This page were made by [[Contributors|Tri Phan]]'''

@@ Line 4: / Line 4: @@
 ==Definition of Continuity==
-   Let <math>c</math> be a real number in the interval <math>(a,b)</math>, and let <math>f</math> be a function whose domain contains the interval<math>(a,b)</math> . The function <math>f</math> is continuous at <math>c</math> when
+   Let <math>c</math> be a real number in the interval <math>(a,b)</math>, and let <math>f</math> be a function whose domain contains the interval <math>(a,b)</math> . The function <math>f</math> is continuous at <math>c</math> when
    these conditions are true.
 . <math>f(c)</math> is defined.
@@ Line 10: / Line 10: @@
 . <math>\lim_{x\to c} f(x)=f(c)</math>
    If <math>f</math> is continuous at every point in the interval <math>(a,b)</math>, then <math>f</math> is continuous on the '''open interval''' <math>(a,b)</math>.
 ==Continuity of piece-wise functions==
 Discuss the continuity of <math>f(x)=\begin{cases}

@@ Line 1: / Line 1: @@
 ==Continuity==
-  Informally, a function is continuous at <math>x=c</math> means that there is no interruption in the graph of <math>f</math> at <math>c</math>.
+Informally, a function is continuous at <math>x=c</math> means that there is no interruption in the graph of <math>f</math> at <math>c</math>.
 ==Definition of Continuity==