Calc 3.4
Transcript of Calc 3.4
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The slope of the slope is how fast the slope is changing.
3.4 Concavity and the 2nd Derivative Test
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We learned how to locate intervals where a function is increasing or decreasing with the first derivative test. Locating intervals where f ’ increases or decreases determines where graph is curving upward or downward.
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To find open intervals where the function is concave upward or downward, we need to find intervals on which f ‘ is increasing or decreasing.
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So to use this test for concavity, we need to locate x-values where f “(x) = 0 or is undefined. These values separate the intervals. Finally, test the sign of the test values in these intervals.
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Ex 1 p. 191 Determining Concavity
Determine the open intervals on which the graph of
is concave upward or downward. 2
3( )
6f x
x
Solution: The function is continuous on the real numbers (no domain problems!) Next, find the second derivative.
2 2'( ) 3( 6) (2 )f x x x 2 2
6
( 6)
x
x
2 2 2 1
2 4
( 6) ( 6) ( 6 )(2)( 6) (2 )''( )
( 6)
x x x xf x
x
2 2 2
2 4
( 6)( 6) ( 6) 4
( 6)
x x x
x
2 13( 6)x
2
2 3
( 6) 3 6
( 6)
x
x
Because f “ (x) = 0 at now we know interval borders
2x
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2
2 3
(18) 2"( )
( 6)
xf x
x
Interval
Test Value
x = -3 x = 0 x =3
Sign of f ”(x)
f ”(-3) > 0
f ”(0) < 0 f ”(3) > 0
Conclusion
Concave up
Concave down
Concave up
( , 2) ( 2, 2) ( 2, )
1
0.5
5
(- 2, .375) ( 2, .375)
f x = 3
x2+6
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Ex 2 p 192 Determining ConcavityDetermine intervals on which the graph of
is concave upward or downward.
2
2
1( )
4
xf x
x
Solution. Differentiate twice, find zeros or undefined values2 2
2 2 2 2
( 4)(2 ) ( 1)(2 ) (2 )( 5)'( )
( 4) ( 4)
x x x x xf x
x x
2 2
10
( 4)
x
x
2 2 2 1
2 4
( 4) ( 10) ( 10 )(2)( 4) (2 )"( )
( 4)
x x x xf x
x
2
2 3
10(3 4)
( 4)
x
x
"( ) 0 can't happen. The function is not continuous at x = 2f x
Use x = -2 and 2 to split intervals
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2
2 3
10(3 4)"( )
( 4)
xf x
x
Intervals (-∞, -2) (-2, 2) (2, ∞)
Test value x = -3 x = 0 x = 3
Sign of f “ f “(-3) > 0 f “(0) < 0 f “(3) > 0
Conclusion Concave up
Concave down
Concave up
4
2
-2
-4
-5 5
concavedown
concave upconcave up
f x = x2+1
x2-4
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Our last problem had two points of inflection, which can be thought of as points where the graph changes from concave upward to concave downward or vice versa.
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Ex 3 p. 193 Points of Inflection
Determine points of inflection and discuss concavity of 4 3( ) 4f x x x
3 2'( ) 4 12f x x x 2"( ) 12 24 12 ( 2)f x x x x x
12 ( 2) 0x x when x=0, 2, so possible points of inflection are when x = 0, or 2
Interval (-∞, 0) (0, 2) (2, ∞)
Test value x =-1 x = 1 x = 3
Sign of f “ f “(-1) > 0 f “(1) < 0 f “(3) > 0
Conclusion Concave up
Concave down
Concave up
Points of inflection happen when x = 0, and 2
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Ex 4. p195 Using the Second Derivative Test to find relative max or mins
Find the relative extrema of f(x) = -3x5 + 5x3
Find critical numbers first (what makes first derivative = 0) 4 2 2 2'( ) 15 15 15 ( 1)f x x x x x
0, 1,1x So critical numbers are
Using f “(x) = -60x3 + 30x, apply 2nd Derivative test.
Point on f(x)
(-1, -2) (1, 2) (0, 0)
Sign of f”(x)
f ” (-1) > 0 f “(1) < 0 f “(0) = 0
Conclusion Concave up so relative min
Concave down so relative max
Test fails
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Looking on either side of x = 0, the first derivative is positive, so at x = 0 is neither a max or min.
2
2
1
-1
-2
-3
rel max at x = 1
test failedat x=0
Rel min at x = -1
f x = -3x5+5x3
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To summarize:
Concavity •look for zeros of f” and see where it is zero or function is not continuous.•Set up intervals with those values•Test intervals – if f”>0 it is concave up. If f” < 0 it is concave down in interval
2nd Derivative test•Look for critical numbers of FIRST derivative•Evaluate SECOND derivative at those values•If f” > 0 then that critical number is relative min•If f” < 0 then that critical number is relative max•If f” = 0, then test fails and you’ll have to look at 1st Derivative test to determine max, min or neither.
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You’ll need to get in your head the 1st der. Test and 2nd der. Test and how they are used!
3.4 Assign. P. 195/ 1-69 every other odd, 79-82