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# Differentiation jan 21, 2014

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Stationary Points,
To determine Maximum / Minimum / Inflection Points
First Derivative Test
Second Derivative Test

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### Differentiation jan 21, 2014

1. 1. Differentiation Chap 9 Objective: How to find Stationary Points & determine their nature (maximum/minimum) riazidan
2. 2. The stationary points of a curve are the points where the gradient is zero e.g. y = x3 − 3x2 − 9x A local maximum x dy =0 dx x A local minimum The word local is usually omitted and the points called maximum and minimum points.
3. 3. e.g.1 Find the coordinates of the stationary points on the curve y = x 3 − 3 x 2 − 9 x y = x3 − 3x2 − 9x Solution: dy ⇒ = 3x2 − 6x − 9 dx dy ⇒ 3 x 2 − 6 x − 9 = 0 ⇒ 3( x 2 − 2 x − 3) = 0 =0 dx 3( x − out + 1 = 0 ⇒ x = 3 Tip: Watch 3)( xfor )common factors or x = −1 x = 3 when finding )stationary points. ⇒ y = ( 3 3 − 3( 3) 2 − 9( 3) = 27 − 27 − 27 = − 27 x = −1 ⇒ y = ( −1) 3 − 3( −1) 2 − 9( −1) = −1 − 3 + 9 = 5 The stationary points are (3, -27) and ( -1, 5)
4. 4. Exercises Find the coordinates of the stationary points of the following functions 2 1. y = x − 4 x + 5 2. y = 2 x 3 + 3 x 2 − 12 x + 1 Solutions: dy 1. = 2x − 4 dx dy = 0 ⇒ 2x − 4 = 0 dx ⇒ x=2 x = 2 ⇒ y = ( 2) 2 − 4( 2) + 5 = 1 Ans: St. pt. is ( 2, 1)
5. 5. y = 2 x 3 + 3 x 2 − 12 x + 1 2. Solution: dy = 6 x 2 + 6 x − 12 dx dy = 0 ⇒ 6( x 2 + x − 2) = 0 ⇒ 6( x − 1)( x + 2) = 0 dx ⇒ x = 1 or x = −2 x = 1 ⇒ y = −6 x = −2 ⇒ y = 2( −2) 3 + 3( −2) 2 − 12( −2) + 1 = 21 Ans: St. pts. are ( 1, −6) and ( −2, 21 )
6. 6. We need to be able to determine the nature of a stationary point ( whether it is a max or a min ). There are several ways of doing this. e.g. On the left of a maximum, the gradient is positive + On the right of a maximum, the gradient is negative −
7. 7. So, for a max the gradients are 0 At the max On the left of On the right of the max the max − + The opposite is true for a minimum − 0 + Calculating the gradients on the left and right of a stationary point tells us whether the point is a max or a min.
8. 8. e.g.2 Find the coordinates of the stationary point of the 2 curve y = x − 4 x + 1 . Is the point a max or min? − − − − − − (1) y = x2 − 4x + 1 Solution: dy ⇒ = 2x − 4 dx dy =0 ⇒ 2x − 4 = 0 ⇒ x = 2 dx y = ( 2) 2 − 4( 2) + 1 ⇒ y = −3 Substitute in (1): dy = 2(1) − 4 = − 2 < 0 On the left of x = 2 e.g. at x = 1, dx dy On the right of x = 2 e.g. at x = 3, = 2( 3) − 4 = 2 > 0 dx + − ⇒ ( 2, − 3) is a min We have 0
9. 9. Another method for determining the nature of a stationary point. e.g.3 Consider y = x 3 + 3 x 2 − 9 x + 10 The gradient function is given by dy = 3x2 + 6x − 9 dx dy dx 3 2 At the max of y = x + 3 x − 9 x + 10 the gradient is 0 but the gradient of the gradient is negative.
10. 10. Another method for determining the nature of a stationary point. e.g.3 Consider y = x 3 + 3 x 2 − 9 x + 10 The gradient function is given by dy = 3x2 + 6x − 9 dx dy dx At the min of y = x 3 + 3 x 2 − 9 x + 10 the gradient of the gradient is positive. d2y The notation for the gradient of the gradient is dx 2 “d 2 y by d x squared”
11. 11. e.g.3 ( continued ) Find the stationary points on the curve y = x 3 + 3 x 2 − 9 x + 10 and distinguish between the max and the min. y = x 3 + 3 x 2 − 9 x + 10 Solution: dy d2y 2 ⇒ = 3x + 6x − 9 ⇒ = 6x + 6 2 dx dx 2 dy 2 d y Stationary points: = 0 ⇒ 3 x + 6 x −is called the 9=0 dx dx 2 nd 2 derivative ⇒ 3( x 2 + 2 x − 3) = 0 ⇒ 3( x + 3)( x − 1) = 0 ⇒ x = −3 or x = 1 We now need to find the y-coordinates of the st. pts.
12. 12. y = x 3 + 3 x 2 − 9 x + 10 x = −3 ⇒ y = ( −3) 3 + 3( −3) 2 − 9( −3) + 10 = 37 x =1 y = 1 + 3 − 9 + 10 = 5 ⇒ To distinguish between max and min we use the 2nd derivative, at the stationary points. d2y 2 = 6x + 6 dx d y = 6( −3) + 6 = −12 < 0 ⇒ max at (−3, 37 ) At x = −3 , 2 dx 2 At x = 1 , d2y dx 2 = 6 + 6 = 12 > 0 ⇒ min at (1, 5)
13. 13. SUMMARY  To find stationary points, solve the equation dy =0 dx  Determine the nature of the stationary points • either by finding the gradients on the left and right of the stationary points + − • ⇒ minimum 0 0 + − ⇒ maximum or by finding the value of the 2nd derivative at the stationary points d2y dx 2 < 0 ⇒ max d2y dx 2 > 0 ⇒ min
14. 14. Exercises Find the coordinates of the stationary points of the following functions, determine the nature of each and sketch the functions. 3 2 3 2 y = x + 3x − 2 Ans. (0, − 2) is a min. 1. (−2 , 2) 2. y = x + 3x − 2 is a max. y = 2 + 3x − x3 Ans. (−1, 0) (1 , 4) is a min. is a max. y = 2 + 3x − x3
15. 15. The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.
16. 16. The stationary points of a curve are the points where the gradient is zero e.g. y = x3 − 3x2 − 9x A local maximum x dy =0 dx x A local minimum The word local is usually omitted and the points called maximum and minimum points.
17. 17. e.g.1 Find the coordinates of the stationary points y = x3 − 3x2 − 9x on the curve Solution: ⇒ dy =0 dx ⇒ y = x3 − 3x2 − 9x dy = 3x2 − 6x − 9 dx 3x2 − 6x − 9 = 0 ⇒ 3( x 2 − 2 x − 3) = 0 3( x − 3)( x + 1) = 0 ⇒ x = 3 or x = −1 x = 3 ⇒ y = ( 3) 3 − 3( 3) 2 − 9( 3) = 27 − 27 − 27 = − 27 x = −1 ⇒ y = ( −1) 3 − 3( −1) 2 − 9( −1) = −1 − 3 + 9 = 5 The stationary points are (3, -27) and ( -1, 5)
18. 18. Determining the nature of a Stationary Point For a max we have On the left of the max + 0 At the max − On the right of the max The opposite is true for a minimum − 0 + Calculating the gradients on the left and right of a stationary point tells us whether the point is a max or a min.
19. 19. Another method for determining the nature of a stationary point. e.g. Consider y y = x 3 + 3 x 2 − 9 x + 10 The gradient function is given by dy = 3x2 + 6x − 9 dx dy dx 3 2 At the max of y = x + 3 x − 9 x + 10 the gradient is 0, but the gradient of the gradient is negative.
20. 20. y = x 3 + 3 x 2 − 9 x + 10 The gradient function is given by dy = 3x2 + 6x − 9 dx dy dx At the min of y = x 3 + 3 x 2 − 9 x + 10 the gradient of the gradient is positive. d2y The notation for the gradient of the gradient is dx 2 “d 2 y by d x squared”
21. 21. The gradient of the gradient is called the 2nd derivative and is written as d2y dx 2
22. 22. e.g. Find the stationary points on the curve 3 y = xand 3distinguish between the max + x 2 − 9 x + 10 and the=min.+ 3 x 2 − 9 x + 10 y x3 Solution: dy d2y 2 ⇒ = 3x + 6x − 9 ⇒ = 6x + 6 2 dx dx dy Stationary points: = 0 ⇒ 3x2 + 6x − 9 = 0 dx ⇒ 3( x 2 + 2 x − 3) = 0 ⇒ 3( x + 3)( x − 1) = 0 ⇒ x = −3 or x = 1 We now need to find the y-coordinates of the st. pts.
23. 23. y = x 3 + 3 x 2 − 9 x + 10 x = −3 ⇒ y = ( −3) 3 + 3( −3) 2 − 9( −3) + 10 = 37 x =1 y = 1 + 3 − 9 + 10 = 5 ⇒ To distinguish between max and min we use the 2nd derivative, d2y 2 = 6x + 6 dx d2y At x = −3 , 2 = 6( −3) + 6 = −12 < 0 ⇒ max at (−3, 37 ) dx At x =1 , d2y dx 2 = 6 + 6 = 12 > 0 ⇒ min at (1, 5)
24. 24. SUMMARY  To find stationary points, solve the equation dy =0 dx  Determine the nature of the stationary points • either by finding the gradients on the left and right of the stationary points 0 − + − ⇒ maximum ⇒ minimum + 0 • or by finding the value of the 2nd derivative at the stationary points d2y dx 2 < 0 ⇒ max d2y dx 2 > 0 ⇒ min