Difference between revisions of "Math 22 Concavity and the Second-Derivative Test"
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|Then we have: <math>f''(-1)=24>0</math>, <math>f''(\frac{1}{2})=-3<0</math> and <math>f''(2)=24>0</math> | |Then we have: <math>f''(-1)=24>0</math>, <math>f''(\frac{1}{2})=-3<0</math> and <math>f''(2)=24>0</math> | ||
|- | |- | ||
− | |'''Step 4''': By the concavity test, <math>f(x)</math> is concave up in <math>( | + | |'''Step 4''': By the concavity test, <math>f(x)</math> is concave up in <math>(-\infty,0)\cup (1,\infty)</math> and <math>f(x)</math> is concave down in <math>(0,1)</math> |
+ | |} | ||
+ | ==Points of Inflection== | ||
+ | If the graph of a continuous function has a tangent line at a point | ||
+ | where its concavity changes from upward to downward (or downward to upward), | ||
+ | then the point is a point of inflection. | ||
+ | |||
+ | If <math>(c,f(c))</math> is a point of inflection of the graph of <math>f</math>, | ||
+ | then either <math>f''(c)=0</math> or <math>f''(c)</math> is undefined. | ||
+ | |||
+ | In '''exercises 1''', at <math>x=0</math>, the concavity changes from concave down to concave up, so <math>(0,f(0))</math> is a point of inflection. | ||
+ | |||
+ | Therefore, <math>(0,0)</math> is a point of inflection | ||
+ | |||
+ | |||
+ | In '''exercises 2''', at <math>x=0</math> and <math>x=1</math>, the concavity changes from concave up to concave down and from concave down to concave up, respectively. So, <math>(0,f(0))</math> and <math>(1,f(1))</math> are points of inflection. | ||
+ | |||
+ | Therefore, <math>(0,10)</math> and <math>(1,9)</math> are points of inflection. | ||
+ | ==The Second-Derivative Test== | ||
+ | Let <math>f'(c)=0</math>, and let <math>f''(x)</math> exist on an open interval containing <math>c</math>, | ||
+ | 1. If <math>f''(c)>0</math>, then <math>f(c)</math> is relative minimum. | ||
+ | 2. If <math>f''(c)<0</math>, then <math>f(c)</math> is relative maximum. | ||
+ | 3. If <math>f''(c)=0</math>, then the test fails. Use the first derivative test. | ||
+ | |||
+ | '''Exercises:''' Find all relative extrema of | ||
+ | |||
+ | '''1)''' <math>f(x)=2x^3+3x^2-5</math> | ||
+ | {| class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | !Solution: | ||
+ | |- | ||
+ | |Notice, <math>f'(x)=6x^2-6x=6x(x-1)=0</math>, then critical numbers are <math>x=0</math> and <math>x=1</math> | ||
+ | |- | ||
+ | |<math>f''(x)=12x-6</math>, then <math>f''(0)=-6<0</math> and <math>f''(1)=6>0</math> | ||
+ | |- | ||
+ | |By the second derivative test, <math>f(0)=-5</math> is relative maximum and <math>f(1)=0</math> is relative minimum | ||
+ | |- | ||
+ | |Therefore, relative maximum: <math>(0,-5)</math> and relative minimum : <math>(1,0)</math> | ||
|} | |} | ||
Latest revision as of 07:29, 31 July 2020
Formal Definition of Concavity
Let be differentiable on an open interval . The graph of is 1. Concave upward on when is increasing on the interval. 2. Concave downward on when is decreasing on the interval.
Test for Concavity
Let be a function whose second derivative exists on an open interval 1. If for all in , then the graph of is concave upward on . 2. If for all in , then the graph of is concave downward on .
Guidelines for Applying the Concavity Test
1. Locate the -values at which or is undefined. 2. Use these -values to determine the test intervals. 3. Determine the sign of at an arbitrary number in each test intervals 4. Apply the concavity test
Exercises: Find the second derivative of and discuss the concavity of its graph.
1)
Solution: |
---|
Step 1: , so |
Step 2: So , so the test intervals are and |
Step 3: Choose for the interval , and for the interval . |
Then we have: and |
Step 4: By the concavity test, is concave up in and is concave down in |
2)
Solution: |
---|
Step 1: , so |
Step 2: So, and , so the test intervals are and |
Step 3: Choose for the interval , for the interval and for the interval . |
Then we have: , and |
Step 4: By the concavity test, is concave up in and is concave down in |
Points of Inflection
If the graph of a continuous function has a tangent line at a point where its concavity changes from upward to downward (or downward to upward), then the point is a point of inflection. If is a point of inflection of the graph of , then either or is undefined.
In exercises 1, at , the concavity changes from concave down to concave up, so is a point of inflection.
Therefore, is a point of inflection
In exercises 2, at and , the concavity changes from concave up to concave down and from concave down to concave up, respectively. So, and are points of inflection.
Therefore, and are points of inflection.
The Second-Derivative Test
Let , and let exist on an open interval containing , 1. If , then is relative minimum. 2. If , then is relative maximum. 3. If , then the test fails. Use the first derivative test.
Exercises: Find all relative extrema of
1)
Solution: |
---|
Notice, , then critical numbers are and |
, then and |
By the second derivative test, is relative maximum and is relative minimum |
Therefore, relative maximum: and relative minimum : |
This page were made by Tri Phan