18 uždavinys
Duota funkcija f (x) = 3x2 + 5x4 - cos(πx).
1. Apskaičiuokite f ' (0).
Sprendimas:
Randame f(x) išvestinę ir į ją įstatome x = 0
( 3* x^2+ 5* x^4-cos( π
* x))′ =
* x))′ = |
|
( 3* x^2+ 5* x^4-cos( π* x))′ = $$(3\cdot x^{2}+5\cdot x^{4}-cos(\pi\cdot x))'$$ =
* x))′ = $$(3\cdot x^{2}+5\cdot x^{4}-cos(\pi\cdot x))'$$ = Paaiškinimas: ( 3* x^2)′+ ( 5* x^4)′- cos( π* x)′ = $$(3\cdot x^{2})'+(5\cdot x^{4})'-cos(\pi\cdot x)'$$ =
* x)′ = $$(3\cdot x^{2})'+(5\cdot x^{4})'-cos(\pi\cdot x)'$$ = Paaiškinimas: 6* x+ ( 5* x^4)′- cos( π* x)′ = $$6\cdot x+(5\cdot x^{4})'-cos(\pi\cdot x)'$$ =
* x)′ = $$6\cdot x+(5\cdot x^{4})'-cos(\pi\cdot x)'$$ = Paaiškinimas: 6* x+ 20* x^3- cos( π* x)′ = $$6\cdot x+20\cdot x^{3}-cos(\pi\cdot x)'$$ =
* x)′ = $$6\cdot x+20\cdot x^{3}-cos(\pi\cdot x)'$$ = Paaiškinimas: 6* x+ 20* x^3+ sin( π* x)* ( π* x)′ = $$6\cdot x+20\cdot x^{3}+sin(\pi\cdot x)\cdot (\pi\cdot x)'$$ =
* x)* ( π* x)′ = $$6\cdot x+20\cdot x^{3}+sin(\pi\cdot x)\cdot (\pi\cdot x)'$$ = Paaiškinimas: 6* x+ 20* x^3+ sin( π* x)* π = $$6\cdot x+20\cdot x^{3}+sin(\pi\cdot x)\cdot \pi$$ =
* x)* π = $$6\cdot x+20\cdot x^{3}+sin(\pi\cdot x)\cdot \pi$$ = Paaiškinimas: 6* 0+ 20* 0^3+ sin( π* 0)* π = $$6\cdot 0+20\cdot 0^{3}+sin(\pi\cdot 0)\cdot \pi$$ =
* 0)* π = $$6\cdot 0+20\cdot 0^{3}+sin(\pi\cdot 0)\cdot \pi$$ = 20* 0^3+ sin( π* 0)* π = $$20\cdot 0^{3}+sin(\pi\cdot 0)\cdot \pi$$ =
* 0)* π = $$20\cdot 0^{3}+sin(\pi\cdot 0)\cdot \pi$$ = sin( π* 0)* π = $$sin(\pi\cdot 0)\cdot \pi$$ =
* 0)* π = $$sin(\pi\cdot 0)\cdot \pi$$ = sin(0)* π = $$sin(0)\cdot \pi$$ =
= $$sin(0)\cdot \pi$$ = 0* π = $$0\cdot \pi$$ =
= $$0\cdot \pi$$ = 0$$0$$
Atsakymas: 0
2. Nustatykite, kokia funkcija yra f ' (x) : lyginė, nelyginė ar nei lyginė, nei nelyginė. Atsakymą pagrįskite.
Sprendimas:
Randame, kam lygi f ' ( - x):
6* x+ 20* x^3+ sin( π* x)* π =
* x)* π = |
|
6* x+ 20* x^3+ sin( π* x)* π = $$6\cdot x+20\cdot x^{3}+sin(\pi\cdot x)\cdot \pi$$ =
* x)* π = $$6\cdot x+20\cdot x^{3}+sin(\pi\cdot x)\cdot \pi$$ = Paaiškinimas: 6* (-x)+ 20* (-x)^3+ sin( π* (-x))* π = $$6\cdot (-x)+20\cdot (-x)^{3}+sin(\pi\cdot (-x))\cdot \pi$$ =
* (-x))* π = $$6\cdot (-x)+20\cdot (-x)^{3}+sin(\pi\cdot (-x))\cdot \pi$$ = - 6* x+ 20* (-x)^3+ sin( π* (-x))* π = $$-6\cdot x+20\cdot (-x)^{3}+sin(\pi\cdot (-x))\cdot \pi$$ =
* (-x))* π = $$-6\cdot x+20\cdot (-x)^{3}+sin(\pi\cdot (-x))\cdot \pi$$ = - 6* x- 20* x^3+ sin( π* (-x))* π = $$-6\cdot x-20\cdot x^{3}+sin(\pi\cdot (-x))\cdot \pi$$ =
* (-x))* π = $$-6\cdot x-20\cdot x^{3}+sin(\pi\cdot (-x))\cdot \pi$$ = - 6* x- 20* x^3+ sin(- x* π)* π = $$-6\cdot x-20\cdot x^{3}+sin(-x\cdot \pi)\cdot \pi$$ =
)* π = $$-6\cdot x-20\cdot x^{3}+sin(-x\cdot \pi)\cdot \pi$$ = - 6* x- 20* x^3- sin( x* π)* π$$-6\cdot x-20\cdot x^{3}-sin(x\cdot \pi)\cdot \pi$$
)* π$$-6\cdot x-20\cdot x^{3}-sin(x\cdot \pi)\cdot \pi$$Matome, kad f ' ( - x) = - f ' (x), todėl funkcija f ' (x) yra nelyginė.
Atsakymas: nelyginė
3. Apskaičiuokite $$f(2)+\int_{0}^{1} (f(x)\cdot d\cdot x)$$.
Sprendimas:
Randame f (2):
3* x^2+ 5* x^4-cos( π* x) =
* x) = |
|
3* x^2+ 5* x^4-cos( π* x) = $$3\cdot x^{2}+5\cdot x^{4}-cos(\pi\cdot x)$$ =
* x) = $$3\cdot x^{2}+5\cdot x^{4}-cos(\pi\cdot x)$$ = Paaiškinimas: 3* 2^2+ 5* 2^4-cos( π* 2) = $$3\cdot 2^{2}+5\cdot 2^{4}-cos(\pi\cdot 2)$$ =
* 2) = $$3\cdot 2^{2}+5\cdot 2^{4}-cos(\pi\cdot 2)$$ = 12+ 5* 2^4-cos( π* 2) = $$12+5\cdot 2^{4}-cos(\pi\cdot 2)$$ =
* 2) = $$12+5\cdot 2^{4}-cos(\pi\cdot 2)$$ = 12+80-cos( π* 2) = $$12+80-cos(\pi\cdot 2)$$ =
* 2) = $$12+80-cos(\pi\cdot 2)$$ = 92-cos( π* 2) = $$92-cos(\pi\cdot 2)$$ =
* 2) = $$92-cos(\pi\cdot 2)$$ = 92-1 = $$92-1$$ =
91$$91$$
$$3\cdot 2^{2}+5\cdot 2^{4}-cos(\pi\cdot 2)$$ = $$$$
$$12+80-cos(\pi\cdot 2)$$ = $$$$
$$92-cos(\pi\cdot 2)$$ = $$$$
$$92-1$$ = $$$$
$$91$$ $$$$
Randame $$\int_{0}^{1} (f(x)\cdot d\cdot x)$$
∫(0;1; 3* x^2+ 5* x^4-cos( π* x)) =
* x)) = |
|
∫(0;1; 3* x^2+ 5* x^4-cos( π* x)) = $$\int_{0}^{1} (3\cdot x^{2}+5\cdot x^{4}-cos(\pi\cdot x))$$ =
* x)) = $$\int_{0}^{1} (3\cdot x^{2}+5\cdot x^{4}-cos(\pi\cdot x))$$ = Paaiškinimas: (0;1; x^3+ x^5-
) = $$(x^{3}+x^{5}-\frac{sin(\pi\cdot x)}{\pi}){\LARGE |}_{0}^{1}$$ =
| sin( π* x) |
| / π |
Paaiškinimas: ( 1^3+ 1^5-
)-( 0^3+ 0^5-
) = $$(1^{3}+1^{5}-\frac{sin(\pi\cdot 1)}{\pi})-(0^{3}+0^{5}-\frac{sin(\pi\cdot 0)}{\pi})$$ =
| sin( π* 1) |
| / π |
| sin( π* 0) |
| / π |
( 1^3+ 1^5-
)-( 0^3+ 0^5-
) = $$(1^{3}+1^{5}-\frac{sin(\pi)}{\pi})-(0^{3}+0^{5}-\frac{sin(\pi\cdot 0)}{\pi})$$ =
| sin(π) |
| / π |
| sin( π* 0) |
| / π |
( 1^3+ 1^5)-( 0^3+ 0^5-
) = $$(1^{3}+1^{5})-(0^{3}+0^{5}-\frac{sin(\pi\cdot 0)}{\pi})$$ =
| sin( π* 0) |
| / π |
( 1^3+ 1^5)-( 0^5-
) = $$(1^{3}+1^{5})-(0^{5}-\frac{sin(\pi\cdot 0)}{\pi})$$ =
| sin( π* 0) |
| / π |
( 1^3+ 1^5)-(-
) = $$(1^{3}+1^{5})-(-\frac{sin(\pi\cdot 0)}{\pi})$$ =
| sin( π* 0) |
| / π |
( 1^3+ 1^5)-() = $$(1^{3}+1^{5})-()$$ =
( 1^3+ 1^5) = $$(1^{3}+1^{5})$$ =
2$$2$$
$$\int_{0}^{1} (3\cdot x^{2}+5\cdot x^{4}-cos(\pi\cdot x))$$ = $$$$
$$(x^{3}+x^{5}-\frac{sin(\pi\cdot x)}{\pi}){\LARGE |}_{0}^{1}$$ = $$$$
$$(1^{3}+1^{5}-\frac{sin(\pi\cdot 1)}{\pi})-(0^{3}+0^{5}-\frac{sin(\pi\cdot 0)}{\pi})$$ = $$$$
$$(1^{3}+1^{5})$$ = $$$$
$$2$$ $$$$
$$f(2)+\int_{0}^{1} (f(x)\cdot d\cdot x) = 91+2 = 93$$
Atsakymas: 93