Math 22 Extrema and First Derivative Test - Revision history

Tphan046: /* Guidelines for Finding Extrema on a Closed Interval */

2020-07-30T17:17:10Z

Guidelines for Finding Extrema on a Closed Interval

Tphan046: /* Guidelines for Finding Extrema on a Closed Interval */

2020-07-30T17:16:25Z

Guidelines for Finding Extrema on a Closed Interval

Tphan046: /* Absolute Extrema */

2020-07-30T17:11:56Z

Absolute Extrema

Tphan046: /* Guidelines for Finding Relative Extrema */

2020-07-30T17:06:34Z

Guidelines for Finding Relative Extrema

Tphan046: /* Guidelines for Finding Relative Extrema */

2020-07-30T17:05:41Z

Guidelines for Finding Relative Extrema

Tphan046: /* Guidelines for Finding Relative Extrema */

2020-07-30T17:00:45Z

Guidelines for Finding Relative Extrema

Tphan046: /* Guidelines for Finding Relative Extrema */

2020-07-30T16:59:48Z

Guidelines for Finding Relative Extrema

Tphan046: /* Guidelines for Finding Relative Extrema */

2020-07-30T16:59:33Z

Guidelines for Finding Relative Extrema

Tphan046: /* Guidelines for Finding Relative Extrema */

2020-07-30T16:58:56Z

Guidelines for Finding Relative Extrema

Tphan046: /* Guidelines for Finding Relative Extrema */

2020-07-30T16:57:17Z

Guidelines for Finding Relative Extrema

@@ Line 91: / Line 91: @@
 |'''Step 2''': <math>f(0)=5</math>
 |-
-|'''Step 3''': <math>f(-3)=-13</math> and <math>f(1)=3/math>
+|'''Step 3''': <math>f(-3)=-13</math> and <math>f(1)=3</math>
 |-
 |'''Step 4''': Absolute Maximum is <math>5</math> at <math>x=0</math>
 |-
-|and absolute minimum is -<math>-13</math> at <math>x=-3</math>
+|and absolute minimum is <math>-13</math> at <math>x=-3</math>
 |}

@@ Line 81: / Line 81: @@
 . The least of these values is the absolute minimum, and the greatest is the maximum.
 [[Math_22| '''Return to Topics Page''']]
 '''This page were made by [[Contributors|Tri Phan]]'''

@@ Line 66: / Line 66: @@
 ==Absolute Extrema==
-The page is under Construction
+  Let <math>f</math> be defined on an interval <math>I</math> containing <math>c</math>.
 [[Math_22| '''Return to Topics Page''']]
 '''This page were made by [[Contributors|Tri Phan]]'''

@@ Line 31: / Line 31: @@
 !Solution: &nbsp;
 |-
-|Step 1: <math>f'(x)=2x+2=2(x+1)=0</math>,
+|'''Step 1''': <math>f'(x)=2x+2=2(x+1)=0</math>,
 |-
-|Step 2: Critical number is <math>x=-1</math>, so the test intervals are <math>(-\infty,-1)</math> and <math>(-1,\infty)</math>
+|'''Step 2''': Critical number is <math>x=-1</math>, so the test intervals are <math>(-\infty,-1)</math> and <math>(-1,\infty)</math>
 |-
-|Step 3: Choose <math>x=-2</math> for the interval <math>(-\infty,-1)</math>, and <math>x=0</math> for the interval <math>(-1,\infty)</math>.
+|'''Step 3''': Choose <math>x=-2</math> for the interval <math>(-\infty,-1)</math>, and <math>x=0</math> for the interval <math>(-1,\infty)</math>.
 |-
 |Then we have: <math>f'(-2)=-2<0</math> and <math>f'(0)=2>0</math>
 |-
-|Step 4: By the first derivative test, <math>f'(x)</math> is negative to the left of <math>x=-1</math> and positive to the right of <math>x=-1</math>, then <math>f(1)=3</math> is a relative minimum
+|'''Step 4''': By the first derivative test, <math>f'(x)</math> is negative to the left of <math>x=-1</math> and positive to the right of <math>x=-1</math>, then <math>f(1)=3</math> is a relative minimum
 |-
 |Therefore, Relative minimum: <math>(1,3)</math>
@@ Line 50: / Line 50: @@
 !Solution: &nbsp;
 |-
-|Step 1: <math>f'(x)=15x^2-20x=5x(3x-4)=0</math>,
+|'''Step 1''': <math>f'(x)=15x^2-20x=5x(3x-4)=0</math>,
 |-
-|Step 2: Critical number is <math>x=0</math> and <math>x=\frac{4}{3}</math>, so the test intervals are <math>(-\infty,0),(0,\frac{4}{3})</math> and <math>(\frac{4}{3},\infty)</math>
+|'''Step 2''': Critical number is <math>x=0</math> and <math>x=\frac{4}{3}</math>, so the test intervals are <math>(-\infty,0),(0,\frac{4}{3})</math> and <math>(\frac{4}{3},\infty)</math>
 |-
-|Step 3: Choose <math>x=-1</math> for the interval <math>(-\infty,0)</math>, <math>x=1</math> for the interval <math>(0,\frac{4}{3})</math> and <math>x=2</math> for the interval <math>(\frac{4}{3},\infty)</math>.
+|'''Step 3''': Choose <math>x=-1</math> for the interval <math>(-\infty,0)</math>, <math>x=1</math> for the interval <math>(0,\frac{4}{3})</math> and <math>x=2</math> for the interval <math>(\frac{4}{3},\infty)</math>.
 |-
 |Then we have: <math>f'(-1)=35>0</math>, <math>f'(1)=-5<0</math> and <math>f'(2)=20>0</math>
 |-
-|Step 4: By the first derivative test, <math>f'(x)</math> is positive to the left of <math>x=0</math> and negative to the right of <math>x=0</math>, then <math>f(0)=3</math> is a relative maximum,
+|'''Step 4''': By the first derivative test, <math>f'(x)</math> is positive to the left of <math>x=0</math> and negative to the right of <math>x=0</math>, then <math>f(0)=3</math> is a relative maximum,
 |-
 | and <math>f'(x)</math> is negative to the left of <math>x=\frac{4}{3}</math> and positive to the right of <math>x=\frac{4}{3}</math>, then <math>f(\frac{4}{3})=\frac{-79}{27}</math> is a relative minimum.

@@ Line 39: / Line 39: @@
 |Then we have: <math>f'(-2)=-2<0</math> and <math>f'(0)=2>0</math>
 |-
-|Step 4: <math>f'(x)</math> is negative to the left of <math>x=-1</math> and positive to the right of <math>x=-1</math>, then <math>f(1)=3</math> is a relative minimum
+|Step 4: By the first derivative test, <math>f'(x)</math> is negative to the left of <math>x=-1</math> and positive to the right of <math>x=-1</math>, then <math>f(1)=3</math> is a relative minimum
 |-
 |Therefore, Relative minimum: <math>(1,3)</math>
@@ Line 56: / Line 56: @@
 |Step 3: Choose <math>x=-1</math> for the interval <math>(-\infty,0)</math>, <math>x=1</math> for the interval <math>(0,\frac{4}{3})</math> and <math>x=2</math> for the interval <math>(\frac{4}{3},\infty)</math>.
 |-
-|Then we have: <math>f'(-2)=-2<0</math> and <math>f'(0)=2>0</math>
+|Then we have: <math>f'(-1)=35>0</math>, <math>f'(1)=-5<0</math> and <math>f'(2)=20>0</math>
 |-
-|Step 4: <math>f'(x)</math> is negative to the left of <math>x=-1</math> and positive to the right of <math>x=-1</math>, then <math>f(1)=3</math> is a relative minimum
+|Step 4: By the first derivative test, <math>f'(x)</math> is positive to the left of <math>x=0</math> and negative to the right of <math>x=0</math>, then <math>f(0)=3</math> is a relative maximum,
 |}

@@ Line 52: / Line 52: @@
 |Step 1: <math>f'(x)=15x^2-20x=5x(3x-4)=0</math>,
 |-
-|Step 2: Critical number is <math>x=0</math> and <math>x=\frac{4}{3}</math>, so the test intervals are <math>(-\infty,0),<math>(0,\frac{4}{3})</math></math> and <math>(\frac{4}{3},\infty)</math>
+|Step 2: Critical number is <math>x=0</math> and <math>x=\frac{4}{3}</math>, so the test intervals are <math>(-\infty,0),(0,\frac{4}{3})</math></math> and <math>(\frac{4}{3},\infty)</math>
 |-
 |Step 3: Choose <math>x=-2</math> for the interval <math>(-\infty,-1)</math>, and <math>x=0</math> for the interval <math>(-1,\infty)</math>.

@@ Line 46: / Line 46: @@
 |}
-'''2)''' <math>f(x)=6x^{\frac{2}{3}}+4x^2+3</math>
+'''2)''' <math>f(x)=5x^3-10x^2+3</math>
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 !Solution: &nbsp;
 |-
-|Step 1: <math>f'(x)=6\frac{2}{3}x^{(\frac{2}{3}-1)}+8x=4x^{\frac{-1}{3}}+8x=4x(x^{\frac{-4}{3}}+2)</math>,
+|Step 1: <math>f'(x)=15x^2-20x=5x(3x-4)=0</math>,
 |-
-|Step 2: Critical number is <math>x=0</math>, so the test intervals are <math>(-\infty,-1)</math> and <math>(-1,\infty)</math>
+|Step 2: Critical number is <math>x=0</math> and <math>x=\frac{4}{3}</math>, so the test intervals are <math>(-\infty,0),<math>(0,\frac{4}{3})</math></math> and <math>(\frac{4}{3},\infty)</math>
 |-
 |Step 3: Choose <math>x=-2</math> for the interval <math>(-\infty,-1)</math>, and <math>x=0</math> for the interval <math>(-1,\infty)</math>.

@@ Line 40: / Line 40: @@
 |-
 |Step 4: <math>f'(x)</math> is negative to the left of <math>x=-1</math> and positive to the right of <math>x=-1</math>, then <math>f(1)=3</math> is a relative minimum
 |-
 |(Note: <math>f(x)</math> in this case is a parabola so our answer makes sense)
@@ Line 48: / Line 50: @@
 !Solution: &nbsp;
 |-
-|<math style="vertical-align: -5px">f(x)=x^{\frac{1}{2}}</math>
+|Step 1: <math>f'(x)=6\frac{2}{3}x^{(\frac{2}{3}-1)}+8x=4x^{\frac{-1}{3}}+8x=4x(x^{\frac{-4}{3}}+2)</math>,
 |-
-|So, <math>f'(x)=\frac{1}{2}x^{\frac{-1}{2}}=\frac{1}{2\sqrt{x}}</math>
+|Then we have: <math>f'(-2)=-2<0</math> and <math>f'(0)=2>0</math>
 |-
-|In this case, we have critical number when <math>f'(x)</math> is undefined, which is when <math>\sqrt{x}=0</math>. So critical number is <math>x=0</math>
+|Step 4: <math>f'(x)</math> is negative to the left of <math>x=-1</math> and positive to the right of <math>x=-1</math>, then <math>f(1)=3</math> is a relative minimum
 |}