Задание 24.2

24.2. Найдите наибольшее и наименьшее значения функции f на указанном промежутке:

1.
f(x) = 1 3x3 − 4x,[0;3];
2.
f(x) = x − 1 − x3 − x2,[−2;0];
3.
f(x) = 2x4 − 8x,[−2;1];
4.
f(x) = x4 4 − 8x2,[−1;2].

Answers

User profile picture
2021-12-09 09:43
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1.
f(x) = 1 3x3 − 4x,[0;3]

f′(x) = x2 − 4

x2 − 4 = 0

x2 = 4

x1 = 2,x2 = −2

2 ∈ [0;3],−2∉[0;3]

f(0) = 1 3 ⋅ 0 − 4 ⋅ 0 = 0

f(2) = −1 3 ⋅ 23 − 4 ⋅ 2 = 8 (1 3 − 1) = 8 ⋅ (−2 3 ) = −51 3

f(3) = 1 3 ⋅ 33 − 4 ⋅ 3 = 32 − 12 = −3

max ⁡ [0;3]f(x) = 0

min ⁡ [0;3]f(x) = −51 3

2.
f(x) = x − 1 − x3 − x2,[−2;0]

f(x) = 1 − 3x2 − 2x

−3x2 − 2x + 1 = 0

D = 4 + 12 = 16

x1 = 6 − 6 = −1

x2 = − 2 − 6 = 1 3

1 3∉[−2;0]

f(−2) = −2 − 1 − (−2)3 − (−2)2 = −2 − 1 + 8 − 4 = −2

f(0) = 0 − 1 − 0 − 0 = −1

max ⁡ [−2;0]f(x) = 1

min ⁡ 2,0]f(x) = −2

3.
f(x) = 2x4 − 8x,[−2;1]

f′(x) = 8x3 − 8

8x3 − 8 = 0

8x3 = 8

x3 = 1

x1 = 1

1 ∈ [−2;1]

f(1) = 2 − 8 = −6

f(−2) = 2 ⋅ (−2)4 − 8 ⋅ (−2) = 32 + 16 = 48

max ⁡ [−2;1]f(x) = 48

min ⁡ [−2;1]f(x) = −6

4.
f(x) = x4 4 − 8x2,[−1;2]

f′(x) = x3 − 16x

x3 − 16x = 0

x1 = 0x2 = 16

x2 = 4,x3 = −4

4,−4∉[−1;2]

f(0) = 0

f(−1) = 1 4 − 8 = −73 4

f(2) = 16 4 − 8 ⋅ 4 = 4 − 32 = −28

max ⁡ [−1;2]f(x) = 0

min ⁡ [−1;2]f(x) = −28

User profile picture
2021-12-09 09:59
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