22 uždavinys24 uždavinys
Duota funkcija f(x) = $$x^{3}-6\cdot x^{2}+8\cdot x+6$$. Tiesė y = $$k\cdot x+b$$ yra funkcijos f (x) grafiko liestinė taške x0 = 3.
1. Apskaičiuokite k ir b reikšmes.
Sprendimas:
f'(x) = $$(x^{3}-6\cdot x^{2}+8\cdot x+6)' = 3\cdot x^{2}-12\cdot x+8$$
k = f'(x0) = f'(3) = $$3\cdot 3^{2}-12\cdot 3+8$$ = $$27-36+8$$ = -1
Liestinės lygtis taške (x0; f(x0)):
y = f(x0)+f'(x0)*(x-x0) = f(3)+f'(3)*(x-3);
f(3) = $$3^{3}-6\cdot 3^{2}+8\cdot 3+6 = 27-54+24+6 = 3$$
$$y = 3-1\cdot (x-3) = -x+6$$
Atsakymas: k = -1; b = 6
2. Apskaičiuokite figūros, kurią riboja funkcijos f (x) grafikas ir jo liestinė taške x0 = 3 plotą.
Sprendimas:
Sulyginame f(x) ir liestinės lygtis:
x^3- 6* x^2+ 8* x+6 =
-x+6
x^3- 6* x^2+ 8* x+6 = -x+6$$x^{3}-6\cdot x^{2}+8\cdot x+6$$ = $$-x+6$$
x^3- 6* x^2+ 8* x+x+6 = 6$$x^{3}-6\cdot x^{2}+8\cdot x+x+6$$ = $$6$$
x^3- 6* x^2+ 8* x+x+6-6 = 0$$x^{3}-6\cdot x^{2}+8\cdot x+x+6-6$$ = $$0$$
x^3- 6* x^2+ 8* x+x+0 = 0$$x^{3}-6\cdot x^{2}+8\cdot x+x+0$$ = $$0$$
x^3- 6* x^2+ 8* x+x = 0$$x^{3}-6\cdot x^{2}+8\cdot x+x$$ = $$0$$
x^3- 6* x^2+ 9* x = 0$$x^{3}-6\cdot x^{2}+9\cdot x$$ = $$0$$
x = 0$$x$$ = $$0$$
x-3 = 0$$x-3$$ = $$0$$
x = 0+3$$x$$ = $$0+3$$
x = 3$$x$$ = $$3$$
Grafikai kertasi taškuose x = 0 ir x = 3.
Norint rasti ieškomą plotą, reikia apskaičiuoti grafiko f(x) ir liestinės lygties y = -x + 6 skirtumo integralą nuo 0 iki 3:
∫(0;3; x^3- 6* x^2+ 8* x+6-(-x+6)) =
∫(0;3; x^3- 6* x^2+ 8* x+6-(-x+6)) = $$\int_{0}^{3} (x^{3}-6\cdot x^{2}+8\cdot x+6-(-x+6))$$ =
∫(0;3; x^3- 6* x^2+ 8* x+6+x-6) = $$\int_{0}^{3} (x^{3}-6\cdot x^{2}+8\cdot x+6+x-6)$$ =
∫(0;3; x^3- 6* x^2+ 8* x+x+6-6) = $$\int_{0}^{3} (x^{3}-6\cdot x^{2}+8\cdot x+x+6-6)$$ =
∫(0;3; x^3- 6* x^2+ 9* x+6-6) = $$\int_{0}^{3} (x^{3}-6\cdot x^{2}+9\cdot x+6-6)$$ =
∫(0;3; x^3- 6* x^2+ 9* x+0) = $$\int_{0}^{3} (x^{3}-6\cdot x^{2}+9\cdot x+0)$$ =
∫(0;3; x^3- 6* x^2+ 9* x) = $$\int_{0}^{3} (x^{3}-6\cdot x^{2}+9\cdot x)$$ =
|(0;3; - 2* x^3+ * x^2) = $$(\frac{x^{4}}{4}-2\cdot x^{3}+\frac{9}{2}\cdot x^{2}){\LARGE |}_{0}^{3}$$ = |(0;3; - 2* x^3+ ) = $$(\frac{x^{4}}{4}-2\cdot x^{3}+\frac{9\cdot x^{2}}{2}){\LARGE |}_{0}^{3}$$ = ( - 2* 3^3+ )-( - 2* 0^3+ ) = $$(\frac{81}{4}-2\cdot 3^{3}+\frac{9\cdot 3^{2}}{2})-(\frac{0^{4}}{4}-2\cdot 0^{3}+\frac{9\cdot 0^{2}}{2})$$ = ( -54+ )-( - 2* 0^3+ ) = $$(\frac{81}{4}-54+\frac{9\cdot 3^{2}}{2})-(\frac{0^{4}}{4}-2\cdot 0^{3}+\frac{9\cdot 0^{2}}{2})$$ = ( -54+ )-( - 2* 0^3+ ) = $$(\frac{81}{4}-54+\frac{81}{2})-(\frac{0^{4}}{4}-2\cdot 0^{3}+\frac{9\cdot 0^{2}}{2})$$ = ( -54+ )-(- 2* 0^3+ ) = $$(\frac{81}{4}-54+\frac{81}{2})-(-2\cdot 0^{3}+\frac{9\cdot 0^{2}}{2})$$ = ( -54+ )-( ) = $$(\frac{81}{4}-54+\frac{81}{2})-(\frac{9\cdot 0^{2}}{2})$$ = ( -54+ )-() = $$(\frac{81}{4}-54+\frac{81}{2})-()$$ = -54+ = $$\frac{81}{4}-54+\frac{81}{2}$$ = ( + -54) = $$(\frac{81}{4}+\frac{81}{2}-54)$$ = ( -54) = $$(\frac{243}{4}-54)$$ = (60.75-54) = $$(60.75-54)$$ =
6.75$$6.75$$
$$\int_{0}^{3} (x^{3}-6\cdot x^{2}+8\cdot x+6-(-x+6))$$ = $$$$
$$\int_{0}^{3} (x^{3}-6\cdot x^{2}+9\cdot x)$$ = $$$$
$$(\frac{x^{4}}{4}-2\cdot x^{3}+\frac{9\cdot x^{2}}{2}){\LARGE |}_{0}^{3}$$ = $$$$
$$(\frac{3^{4}}{4}-2\cdot 3^{3}+\frac{9\cdot 3^{2}}{2})-(\frac{0^{4}}{4}-2\cdot 0^{3}+\frac{9\cdot 0^{2}}{2})$$ = $$$$
$$\frac{81}{4}-54+\frac{81}{2}$$ = $$$$
$$(\frac{243}{4}-54)$$ = $$$$
∫(0;3;x^3-6*x^2+8*x+6-(-x+6)) =
|(0;3;x^4/4-2*x^3+9*x^2/2) =
(3^4/4-2*3^3+9*3^2/2)-(0^4/4-2*0^3+9*0^2/2) =
Atsakymas: 6.75
22 uždavinys24 uždavinys