Higher_Mathematics_Part_1
.pdf2) Points of intersection with the coordinate axis: if x = 0 , then y = 0 ; if y = 0 , then x = 0 or x = 2 . So, the curve passes through the points (0; 0) and (2; 0).
3)Function is neither even nor odd.
4)The points of discontinuity and vertical asymptotes do not exist.
5) Let’s find the derivative y′ = |
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3x 2 − 4x |
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x(x −4 / 3) |
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3 3 (x3 − 2x 2 )2 |
3 x4 (x −2)2 |
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Critical points are |
x = 0 ; 4/3; 2. These points break the numerical axis into |
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four intervals (−∞;0) , |
(0; 4 / 3) , (4 / 3; 2) , (2; ∞) . On each of these intervals the |
derivative y′ has constant sign, so: if x (−∞;0) , then y′ > 0 and the function increases; if x (0; 4) , then y′ < 0 and the function decreases; if x ( 4 ;2) (2; ∞) ,
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then y′ > 0 and the function |
increases. During transition through the point |
x = 0 derivative changes the sign from plus to minus, so, x = 0 is the point of
maximum, and ymax = y(0) = 0 . |
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x = 4 / 3 derivative |
changes the |
sign |
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During transition through the point |
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from minus to plus, so, at this point it is minimum: |
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ymin = y(4 3) = 3 (4 3)2 (4 3 − 2) |
= − 3 32 27 = − 23 4 |
>> – 1.1. |
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During transition through the point |
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x = 2 |
derivative does not change the |
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sign, so, this point is not the point of extremum. |
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6) Let’s find the second derivative |
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x(x − 4 3) |
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x − 4 3 |
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y′′ = |
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3 |
x |
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(x − |
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x(x − 2) |
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3 x(x − 2)2 − (x − 4 3) (x − 2)2 + 2x(x − 2) |
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33 x2 (x − 2)4 |
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3 x2 (x − 2)4 |
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9x(x − 2)2 − (3x − 4)(x − 2)(3x − 2) = − 8 |
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(x − 2) |
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9 |
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(x − 2) |
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9 3 |
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(x − |
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2) |
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Second derivative doesn’t exist if |
x = 0 or |
x = 2 . So, points x = 0; 2 |
are |
the critical points of the second type. Let’s consider the intervals (−∞; 0) , (0; 2) and (2; ∞) . On the intervals (−∞; 0) and (0; 2) y′′ > 0 and the curve is concave; if x (2; ∞) , then y′′ < 0 and the curve is convex. The inflection point has the coordinates (2; 0).
261
7) Let’s find the inclined asymptote:
k = lim |
f (x) |
3 |
x3 − 3x2 |
= 1, |
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x |
= lim |
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x→∞ |
x→∞ |
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b = lim ( f (x) − kx) = lim (3 |
x3 − 2x 2 |
− x) = |
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x→∞ |
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x→∞ |
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= lim |
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x3 − 2x2 − x3 |
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= − |
2 |
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3 |
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x→∞ 3 (x3 − 2x2 )2 + x 3 x3 − 2x2 + x2 |
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In such a way, straight line y = x −2 / 3 is inclined asymptote of our curve.
Other asymptotes don’t exist.
8) Taking into account the spent investigations, we can draw the graph of our function (Fig.3.35).
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4/3
О |
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2 |
х |
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Fig. 3.35
Micromodule 22
CLASS AND HOME ASSIGNMENTS
Let’s find the intervals of increasing and decreasing of the functions:
1. |
y = 6−3x2 −x3 . |
2. |
y = x4 −2x2 . |
3. y = x ln x . |
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4. |
y = x2 e−x . |
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5. |
y = |
x2 +2x |
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x −1 |
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Investigate the function on extremum: |
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6. |
y = x3 −9x2 +15x −10 . |
7. |
y = x5 −5x4 +5x3 +5 . |
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8. |
y = (x −1)2 (x −2)2 . |
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9. |
y = x(x −1)2 (x +1)3 . |
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10. |
y = |
1−x + x2 |
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11. |
y = 3x + |
1 |
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12. y = |
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2x |
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1+ x |
−x2 |
x3 |
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+ x2 |
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1 |
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262
Let’s find maximum and minimum values of the functions:
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2 |
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13. y = xe−x . |
14. y = x −ln(1+ x) . |
15. y = |
ln x |
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x |
Investigate the behavior of the function in neighbourhood of the giving points with the help of the derivatives of higher order:
16.y = 6ex −x3 −3x2 −6x −5 , x0 = 0 .
17.y = x sin x −x2 , x0 = 0 .
Let’s find the intervals of the concavity and convexity, and the inflection points of the curves.
18. |
y = x2 |
− 2x + 1 . |
19. |
y = x3 |
− 1 . |
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20. |
y = x3 |
− 3x2 + 9x + 6 . |
21. |
y = x2 |
ln x . |
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22. |
y = e− x2 |
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24. y = ln x + 2x2 . |
Let’s find the asymptotes of the curves:
25. y = |
6x4 +3x3 |
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1 |
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27. y = x e2 / x +1. |
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5x3 +1 |
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x2 −3x +2 |
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Investigate the functions and sketch the graphs:
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y = |
x3 +4 |
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29. |
y =(2x+3)e−2/(x+1) . 30. |
y = 3 (x +3)x2 . |
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x2 |
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3x4 +1 |
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31. |
y = arctg(sin x) . |
32. |
y = |
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33. |
y = 3ln |
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−1 . |
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x3 |
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−3 |
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ex+1 |
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34. |
y = |
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35. |
y = |
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4x2 |
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36. |
y = |
x3 |
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x +1 |
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3+ x2 |
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x2 − |
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ln x |
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37. |
y = 3 1−x2 . 38. |
y = x2 + |
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2 |
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39. y =16x(x −1)3 . 40. y = |
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Answers |
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1. |
(−∞;−2) і (0; ∞) – |
decreases; |
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(−2;0) – increases. 2. |
(−∞;−1) and |
(0;1) – |
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decreases; |
(−1;0) and (1; ∞) – increases. 3. |
(0;1/ e) – decreases; |
(1/ e; ∞) – increases. |
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4. (−∞;0) and (2; ∞) – decreases; (0;2) – increases. 5. |
(−∞;1− |
3) and (1+ |
3; ∞) – |
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increases; |
(1− 3;1) and |
(1;1+ |
3) – |
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decreases. 6. |
x = 2 – |
maximum, |
x = 3 – |
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minimum. 7. |
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x = 1– maximum, |
x = 3 – minimum. 8. |
x = 1;2 – minimum, |
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x = 3 / 2 – |
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263 |
maximum. |
9. x = −4 / 3;1– minimum, x = 1/ 2 – maximum. 10. x = 1/ 2 – minimum. |
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x = −1 – maximum, x =1– minimum. 12. x = 1 – maximum. 13. ymax = y(1) = 1/ e . |
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14. |
ymin = y(0) = 0 . |
15. |
ymin |
= y(1) = 0 , ymax = y(e2 ) = 4 / e2 .16. |
Minimum. |
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17. |
Maximum. 20. |
(−∞;1) – |
convex, |
(1; ∞) – |
concave, |
(1;13) is inflection point. |
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21. |
(0; e−3 / 2 ) – convex, (e−3 / 2 ; ∞) – concave, (e−3 / 2 ;−3 /(2e3 )) is inflection point. |
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22. |
(−∞;−1/ |
2) (1/ |
2; ∞) – concave, (−1/ |
2;1/ |
2) – convex, (±1/ |
2;e−1/ 2 ) |
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are inflection points. 24. |
(0;1/ 2) – convex, |
(1/ 2; ∞) – concave, (1/ 2;1/ 2 − ln 2) – point |
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of |
bend. 25. |
6x − 5y + 3 = 0 – inclined asymptote, |
x = −1/ 3 5 |
is a vertical asymptote. |
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26. |
x = 1;2 |
are vertical asymptotes, y = 0 |
is a |
horizontal |
asymptote. |
27. |
y = x + 3 . |
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28. |
y = x , |
x = 0 are |
asymptotes; (−∞;0) (2; ∞) – |
increases, |
(0;2) – |
decreases; |
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xmin = 2 , |
ymin = 3 ; (−∞;0) (0; ∞) – concave. 29. Designated everywhere, except |
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x = −1; extremum |
doesn’t |
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function increases; |
(−2;−e2 ) – inflection point; |
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x = −1– asymptote. 32. Odd; (−∞;−1) (1; ∞) – increases, |
(−1;0) (0;1) – decreases; |
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ymax = −4 |
if x = −1, |
ymin |
= 4 |
if x = 1 ; y = 3x , |
x = 0 – asymptotes; |
(−∞;0) – |
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convex, (0; ∞) – concave. 34. Designated |
everywhere, except |
x = −1 ; |
ymin = e |
if |
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x = 0 ; |
x = −1– |
asymptote; |
(−∞;−1) – |
concave down, |
(−1; ∞) – |
concave |
up. |
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35. |
Even; |
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ymax = 0 |
if x = 0 ; |
y = 4 – |
asymptote; |
(±1; 1) |
is inflection point; |
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(−∞;−1) (1; ∞) – |
convex, |
(−1;1) – |
concave. |
39. |
ymin |
= −27 /16 |
if |
x = 1/ 4 ; |
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(−∞;1/ 2) (1; ∞) – concave, (1/ 2;1) – convex; |
(1/ 2;−1) , |
(1;0) are inflection points; |
asymptotes don’t exist.
Micromodule 22
SELF-TEST ASSIGNMENTS
22.1. Find the intervals of increasing and decreasing of the functions:
22.1.1. y = x2 − ln x2 . |
22.1.2. y = |
x − 1(x − 2) . |
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22.1.3. y = (x − 1)2 (x + 1)3 . |
22.1.4. y = arcsin(1+ x) . |
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22.1.5. y = |
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22.1.6. y = |
x −1 |
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+ 100 |
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x + 24 |
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22.1.7. y = |
x2 |
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22.1.8. y = ln x − arctgx . |
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2x |
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22.1.9. y = |
x5 |
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22.1.10. y = x4 ln x . |
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3x |
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Навчальне видання
ДЕНИСЮК Володимир Петрович ГРИШИНА Людмила Іллівна КАРУПУ Олена Вальтерівна ОЛЕШКО Тетяна Анатоліївна ПАХНЕНКО Валерія Валеріївна РЕПЕТА Віктор Кузьмич
ВИЩА МАТЕМАТИКА
У чотирьох частинах
Частина 1
Навчальний посібник
(Англійською мовою)
В авторській редакції
Художник обкладинки Т. Зябліцева
Верстка О. Іваненко
Підп. до друку 22.12.09. Формат 60·84/16. Папір офс. Офс. друк. Ум. друк. арк. 15,81. Обл.-вид. арк. 17,0.
Тираж 300 пр. Замовлення №
Видавництво Національного авіаційного університету «НАУ-друк» 03680, Київ-58, просп. Космонавта Комарова, 1
Свідоцтво про внесення до Державного реєстру ДК № 977 від 05.07.2002
Тел. (044) 406-78-28. Тел./факс: (044) 406-71-33 E-mail: publish@nau.edu.ua