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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;

x^⁴

/ 4

- 2* x^³+

/ 2

* x^²) = $(\frac{x^{4}}{4}-2\cdot x^{3}+\frac{9}{2}\cdot x^{2}){\LARGE |}_{0}^{3}$ =

|(_0;^3;

x^⁴

/ 4

- 2* x^³+

9* x^²

/ 2

) = $(\frac{x^{4}}{4}-2\cdot x^{3}+\frac{9\cdot x^{2}}{2}){\LARGE |}_{0}^{3}$ =

(

/ 4

- 2* 3^³+

9* 3^²

/ 2

)-(

0^⁴

/ 4

- 2* 0^³+

9* 0^²

/ 2

) = $(\frac{81}{4}-2\cdot 3^{3}+\frac{9\cdot 3^{2}}{2})-(\frac{0^{4}}{4}-2\cdot 0^{3}+\frac{9\cdot 0^{2}}{2})$ =

(

/ 4

-54+

9* 3^²

/ 2

)-(

0^⁴

/ 4

- 2* 0^³+

9* 0^²

/ 2

) = $(\frac{81}{4}-54+\frac{9\cdot 3^{2}}{2})-(\frac{0^{4}}{4}-2\cdot 0^{3}+\frac{9\cdot 0^{2}}{2})$ =

(

/ 4

-54+

/ 2

)-(

0^⁴

/ 4

- 2* 0^³+

9* 0^²

/ 2

) = $(\frac{81}{4}-54+\frac{81}{2})-(\frac{0^{4}}{4}-2\cdot 0^{3}+\frac{9\cdot 0^{2}}{2})$ =

(

/ 4

-54+

/ 2

)-(- 2* 0^³+

9* 0^²

/ 2

) = $(\frac{81}{4}-54+\frac{81}{2})-(-2\cdot 0^{3}+\frac{9\cdot 0^{2}}{2})$ =

(

/ 4

-54+

/ 2

)-(

9* 0^²

/ 2

) = $(\frac{81}{4}-54+\frac{81}{2})-(\frac{9\cdot 0^{2}}{2})$ =

(

/ 4

-54+

/ 2

)-() = $(\frac{81}{4}-54+\frac{81}{2})-()$ =

/ 4

-54+

/ 2

= $\frac{81}{4}-54+\frac{81}{2}$ =

(

/ 4

/ 2

-54) = $(\frac{81}{4}+\frac{81}{2}-54)$ =

(

243

/ 4

-54) = $(\frac{243}{4}-54)$ =

(60.75-54) = $(60.75-54)$ =

6.75 $6.75$

$\int_{0}^{3} (x^{3}-6\cdot x^{2}+8\cdot x+6-(-x+6))$ =

$\int_{0}^{3} (x^{3}-6\cdot x^{2}+9\cdot x)$ =

$(\frac{x^{4}}{4}-2\cdot x^{3}+\frac{9\cdot x^{2}}{2}){\LARGE |}_{0}^{3}$ =

$(\frac{3^{4}}{4}-2\cdot 3^{3}+\frac{9\cdot 3^{2}}{2})-(\frac{0^{4}}{4}-2\cdot 0^{3}+\frac{9\cdot 0^{2}}{2})$ =

$\frac{81}{4}-54+\frac{81}{2}$ =

$(\frac{243}{4}-54)$ =

$(60.75-54)$ =

$6.75$

Atsakymas: 6.75

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