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Popular Trigonometría >

6cosh^2(x)+4sinh(x)=7

Solución

6 cosh^{2} (x) + 4 sinh (x) = 7

Solución

x = ln (1.21230 \dots), x = ln (0.45880 \dots)

Grados

x = 11.03066 \dots^{\circ}, x = - 44.64062 \dots^{\circ}

Pasos de solución

6 cosh^{2} (x) + 4 sinh (x) = 7

Re-escribir usando identidades trigonométricas

6 cosh^{2} (x) + 4 sinh (x) = 7

Utilizar la identidad hiperbólica:

sinh (x) = \frac{e ^{x} - e ^{- x}}{2}

6 cosh^{2} (x) + 4 \cdot \frac{e ^{x} - e ^{- x}}{2} = 7

Utilizar la identidad hiperbólica:

cosh (x) = \frac{e ^{x} + e ^{- x}}{2}

6 (\frac{e ^{x} + e ^{- x}}{2})^{2} + 4 \cdot \frac{e ^{x} - e ^{- x}}{2} = 7

6 (\frac{e ^{x} + e ^{- x}}{2})^{2} + 4 \cdot \frac{e ^{x} - e ^{- x}}{2} = 7

6 (\frac{e ^{x} + e ^{- x}}{2})^{2} + 4 \cdot \frac{e ^{x} - e ^{- x}}{2} = 7 : x = ln (1.21230 \dots), x = ln (0.45880 \dots)

6 (\frac{e ^{x} + e ^{- x}}{2})^{2} + 4 \cdot \frac{e ^{x} - e ^{- x}}{2} = 7

Aplicar las leyes de los exponentes

6 (\frac{e ^{x} + e ^{- x}}{2})^{2} + 4 \cdot \frac{e ^{x} - e ^{- x}}{2} = 7

Aplicar las leyes de los exponentes:

a^{b c} = (a^{b})^{c}

e^{- x} = (e^{x})^{- 1}

6 (\frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2})^{2} + 4 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} = 7

6 (\frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2})^{2} + 4 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} = 7

Re escribir la ecuación con

e^{x} = u

6 (\frac{u + ( u ) ^{- 1}}{2})^{2} + 4 \cdot \frac{u - ( u ) ^{- 1}}{2} = 7

Resolver

6 (\frac{u + u ^{- 1}}{2})^{2} + 4 \cdot \frac{u - u ^{- 1}}{2} = 7 : u \approx 1.21230 \dots, u \approx 0.45880 \dots, u \approx - 0.82487 \dots, u \approx - 2.17956 \dots

6 (\frac{u + u ^{- 1}}{2})^{2} + 4 \cdot \frac{u - u ^{- 1}}{2} = 7

Simplificar

\frac{3 ( u ^{2} + 1 ) ^{2}}{2 u ^{2}} + \frac{2 ( u ^{2} - 1 )}{u} = 7

Multiplicar por el mínimo común múltiplo

\frac{3 ( u ^{2} + 1 ) ^{2}}{2 u ^{2}} + \frac{2 ( u ^{2} - 1 )}{u} = 7

Encontrar el mínimo común múltiplo de

2 u^{2}, u : 2 u^{2}

2 u^{2}, u

Mínimo común múltiplo (MCM)

Calcular una expresión que este compuesta de factores que aparezcan tanto en

2 u^{2}

u

= 2 u^{2}

Multiplicar por el mínimo común múltiplo=

2 u^{2}

\frac{3 ( u ^{2} + 1 ) ^{2}}{2 u ^{2}} \cdot 2 u^{2} + \frac{2 ( u ^{2} - 1 )}{u} \cdot 2 u^{2} = 7 \cdot 2 u^{2}

Simplificar

\frac{3 ( u ^{2} + 1 ) ^{2}}{2 u ^{2}} \cdot 2 u^{2} + \frac{2 ( u ^{2} - 1 )}{u} \cdot 2 u^{2} = 7 \cdot 2 u^{2}

Simplificar

\frac{3 ( u ^{2} + 1 ) ^{2}}{2 u ^{2}} \cdot 2 u^{2} : 3 (u^{2} + 1)^{2}

\frac{3 ( u ^{2} + 1 ) ^{2}}{2 u ^{2}} \cdot 2 u^{2}

Multiplicar fracciones:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 ( u ^{2} + 1 ) ^{2} \cdot 2 u ^{2}}{2 u ^{2}}

Eliminar los terminos comunes:

2

= \frac{3 ( u ^{2} + 1 ) ^{2} u ^{2}}{u ^{2}}

Eliminar los terminos comunes:

u^{2}

= 3 (u^{2} + 1)^{2}

Simplificar

\frac{2 ( u ^{2} - 1 )}{u} \cdot 2 u^{2} : 4 u (u^{2} - 1)

\frac{2 ( u ^{2} - 1 )}{u} \cdot 2 u^{2}

Multiplicar fracciones:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 ( u ^{2} - 1 ) \cdot 2 u ^{2}}{u}

Multiplicar los numeros:

2 \cdot 2 = 4

= \frac{4 u ^{2} ( u ^{2} - 1 )}{u}

Eliminar los terminos comunes:

u

= 4 u (u^{2} - 1)

Simplificar

7 \cdot 2 u^{2} : 14 u^{2}

7 \cdot 2 u^{2}

Multiplicar los numeros:

7 \cdot 2 = 14

= 14 u^{2}

3 (u^{2} + 1)^{2} + 4 u (u^{2} - 1) = 14 u^{2}

3 (u^{2} + 1)^{2} + 4 u (u^{2} - 1) = 14 u^{2}

3 (u^{2} + 1)^{2} + 4 u (u^{2} - 1) = 14 u^{2}

Resolver

3 (u^{2} + 1)^{2} + 4 u (u^{2} - 1) = 14 u^{2} : u \approx 1.21230 \dots, u \approx 0.45880 \dots, u \approx - 0.82487 \dots, u \approx - 2.17956 \dots

3 (u^{2} + 1)^{2} + 4 u (u^{2} - 1) = 14 u^{2}

Desarrollar

3 (u^{2} + 1)^{2} + 4 u (u^{2} - 1) : 3 u^{4} + 6 u^{2} + 3 + 4 u^{3} - 4 u

3 (u^{2} + 1)^{2} + 4 u (u^{2} - 1)

(u^{2} + 1)^{2} = u^{4} + 2 u^{2} + 1

(u^{2} + 1)^{2}

Aplicar la formula del binomio al cuadrado:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = u^{2}, b = 1

= (u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2}

Simplificar

(u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2} : u^{4} + 2 u^{2} + 1

(u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2}

Aplicar la regla

1^{a} = 1

1^{2} = 1

= (u^{2})^{2} + 2 \cdot 1 \cdot u^{2} + 1

(u^{2})^{2} = u^{4}

(u^{2})^{2}

Aplicar las leyes de los exponentes:

(a^{b})^{c} = a^{b c}

= u^{2 \cdot 2}

Multiplicar los numeros:

2 \cdot 2 = 4

= u^{4}

2 u^{2} \cdot 1 = 2 u^{2}

2 u^{2} \cdot 1

Multiplicar los numeros:

2 \cdot 1 = 2

= 2 u^{2}

= u^{4} + 2 u^{2} + 1

= u^{4} + 2 u^{2} + 1

= 3 (u^{4} + 2 u^{2} + 1) + 4 u (u^{2} - 1)

Expandir

3 (u^{4} + 2 u^{2} + 1) : 3 u^{4} + 6 u^{2} + 3

3 (u^{4} + 2 u^{2} + 1)

Aplicar la siguiente regla de productos notables

= 3 u^{4} + 3 \cdot 2 u^{2} + 3 \cdot 1

Simplificar

3 u^{4} + 3 \cdot 2 u^{2} + 3 \cdot 1 : 3 u^{4} + 6 u^{2} + 3

3 u^{4} + 3 \cdot 2 u^{2} + 3 \cdot 1

Multiplicar los numeros:

3 \cdot 2 = 6

= 3 u^{4} + 6 u^{2} + 3 \cdot 1

Multiplicar los numeros:

3 \cdot 1 = 3

= 3 u^{4} + 6 u^{2} + 3

= 3 u^{4} + 6 u^{2} + 3

= 3 u^{4} + 6 u^{2} + 3 + 4 u (u^{2} - 1)

Expandir

4 u (u^{2} - 1) : 4 u^{3} - 4 u

4 u (u^{2} - 1)

Poner los parentesis utilizando:

a (b - c) = ab - a c

a = 4 u, b = u^{2}, c = 1

= 4 u u^{2} - 4 u \cdot 1

= 4 u^{2} u - 4 \cdot 1 \cdot u

Simplificar

4 u^{2} u - 4 \cdot 1 \cdot u : 4 u^{3} - 4 u

4 u^{2} u - 4 \cdot 1 \cdot u

4 u^{2} u = 4 u^{3}

4 u^{2} u

Aplicar las leyes de los exponentes:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u = u^{2 + 1}

= 4 u^{2 + 1}

Sumar:

2 + 1 = 3

= 4 u^{3}

4 \cdot 1 \cdot u = 4 u

4 \cdot 1 \cdot u

Multiplicar los numeros:

4 \cdot 1 = 4

= 4 u

= 4 u^{3} - 4 u

= 4 u^{3} - 4 u

= 3 u^{4} + 6 u^{2} + 3 + 4 u^{3} - 4 u

3 u^{4} + 6 u^{2} + 3 + 4 u^{3} - 4 u = 14 u^{2}

Desplace

14 u^{2}

a la izquierda

3 u^{4} + 6 u^{2} + 3 + 4 u^{3} - 4 u = 14 u^{2}

Restar

14 u^{2}

de ambos lados

3 u^{4} + 6 u^{2} + 3 + 4 u^{3} - 4 u - 14 u^{2} = 14 u^{2} - 14 u^{2}

Simplificar

3 u^{4} + 4 u^{3} - 8 u^{2} - 4 u + 3 = 0

3 u^{4} + 4 u^{3} - 8 u^{2} - 4 u + 3 = 0

Encontrar una solución para

3 u^{4} + 4 u^{3} - 8 u^{2} - 4 u + 3 = 0

utilizando el método de Newton-Raphson:

u \approx 1.21230 \dots

3 u^{4} + 4 u^{3} - 8 u^{2} - 4 u + 3 = 0

Definición del método de Newton-Raphson

f (u) = 3 u^{4} + 4 u^{3} - 8 u^{2} - 4 u + 3

Hallar

f^{^{'}} (u) : 12 u^{3} + 12 u^{2} - 16 u - 4

\frac{d}{d u} (3 u^{4} + 4 u^{3} - 8 u^{2} - 4 u + 3)

Aplicar la regla de la suma/diferencia:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (3 u^{4}) + \frac{d}{d u} (4 u^{3}) - \frac{d}{d u} (8 u^{2}) - \frac{d}{d u} (4 u) + \frac{d}{d u} (3)

\frac{d}{d u} (3 u^{4}) = 12 u^{3}

\frac{d}{d u} (3 u^{4})

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d u} (u^{4})

Aplicar la regla de la potencia:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 \cdot 4 u^{4 - 1}

Simplificar

= 12 u^{3}

\frac{d}{d u} (4 u^{3}) = 12 u^{2}

\frac{d}{d u} (4 u^{3})

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{d}{d u} (u^{3})

Aplicar la regla de la potencia:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 4 \cdot 3 u^{3 - 1}

Simplificar

= 12 u^{2}

\frac{d}{d u} (8 u^{2}) = 16 u

\frac{d}{d u} (8 u^{2})

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 8 \frac{d}{d u} (u^{2})

Aplicar la regla de la potencia:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 8 \cdot 2 u^{2 - 1}

Simplificar

= 16 u

\frac{d}{d u} (4 u) = 4

\frac{d}{d u} (4 u)

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{d u}{d u}

Aplicar la regla de derivación:

\frac{d u}{d u} = 1

= 4 \cdot 1

Simplificar

= 4

\frac{d}{d u} (3) = 0

\frac{d}{d u} (3)

Derivada de una constante:

\frac{d}{d x} (a) = 0

= 0

= 12 u^{3} + 12 u^{2} - 16 u - 4 + 0

Simplificar

= 12 u^{3} + 12 u^{2} - 16 u - 4

Sea

u_{0} = 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 1.5 : Δ u_{1} = 0.5

f (u_{0}) = 3 \cdot 1^{4} + 4 \cdot 1^{3} - 8 \cdot 1^{2} - 4 \cdot 1 + 3 = - 2

f^{^{'}} (u_{0}) = 12 \cdot 1^{3} + 12 \cdot 1^{2} - 16 \cdot 1 - 4 = 4

u_{1} = 1.5

Δ u_{1} = ∣ 1.5 - 1 ∣ = 0.5

Δ u_{1} = 0.5

u_{2} = 1.30537 \dots : Δ u_{2} = 0.19462 \dots

f (u_{1}) = 3 \cdot 1. 5^{4} + 4 \cdot 1. 5^{3} - 8 \cdot 1. 5^{2} - 4 \cdot 1.5 + 3 = 7.6875

f^{^{'}} (u_{1}) = 12 \cdot 1. 5^{3} + 12 \cdot 1. 5^{2} - 16 \cdot 1.5 - 4 = 39.5

u_{2} = 1.30537 \dots

Δ u_{2} = ∣ 1.30537 \dots - 1.5 ∣ = 0.19462 \dots

Δ u_{2} = 0.19462 \dots

u_{3} = 1.22652 \dots : Δ u_{3} = 0.07885 \dots

f (u_{2}) = 3 \cdot 1.30537 \dots^{4} + 4 \cdot 1.30537 \dots^{3} - 8 \cdot 1.30537 \dots^{2} - 4 \cdot 1.30537 \dots + 3 = 1.75491 \dots

f^{^{'}} (u_{2}) = 12 \cdot 1.30537 \dots^{3} + 12 \cdot 1.30537 \dots^{2} - 16 \cdot 1.30537 \dots - 4 = 22.25477 \dots

u_{3} = 1.22652 \dots

Δ u_{3} = ∣ 1.22652 \dots - 1.30537 \dots ∣ = 0.07885 \dots

Δ u_{3} = 0.07885 \dots

u_{4} = 1.21271 \dots : Δ u_{4} = 0.01381 \dots

f (u_{3}) = 3 \cdot 1.22652 \dots^{4} + 4 \cdot 1.22652 \dots^{3} - 8 \cdot 1.22652 \dots^{2} - 4 \cdot 1.22652 \dots + 3 = 0.22886 \dots

f^{^{'}} (u_{3}) = 12 \cdot 1.22652 \dots^{3} + 12 \cdot 1.22652 \dots^{2} - 16 \cdot 1.22652 \dots - 4 = 16.56956 \dots

u_{4} = 1.21271 \dots

Δ u_{4} = ∣ 1.21271 \dots - 1.22652 \dots ∣ = 0.01381 \dots

Δ u_{4} = 0.01381 \dots

u_{5} = 1.21230 \dots : Δ u_{5} = 0.00040 \dots

f (u_{4}) = 3 \cdot 1.21271 \dots^{4} + 4 \cdot 1.21271 \dots^{3} - 8 \cdot 1.21271 \dots^{2} - 4 \cdot 1.21271 \dots + 3 = 0.00639 \dots

f^{^{'}} (u_{4}) = 12 \cdot 1.21271 \dots^{3} + 12 \cdot 1.21271 \dots^{2} - 16 \cdot 1.21271 \dots - 4 = 15.64663 \dots

u_{5} = 1.21230 \dots

Δ u_{5} = ∣ 1.21230 \dots - 1.21271 \dots ∣ = 0.00040 \dots

Δ u_{5} = 0.00040 \dots

u_{6} = 1.21230 \dots : Δ u_{6} = 3.5348 E - 7

f (u_{5}) = 3 \cdot 1.21230 \dots^{4} + 4 \cdot 1.21230 \dots^{3} - 8 \cdot 1.21230 \dots^{2} - 4 \cdot 1.21230 \dots + 3 = 5.52123 E - 6

f^{^{'}} (u_{5}) = 12 \cdot 1.21230 \dots^{3} + 12 \cdot 1.21230 \dots^{2} - 16 \cdot 1.21230 \dots - 4 = 15.61963 \dots

u_{6} = 1.21230 \dots

Δ u_{6} = ∣ 1.21230 \dots - 1.21230 \dots ∣ = 3.5348 E - 7

Δ u_{6} = 3.5348 E - 7

u \approx 1.21230 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{3 u ^{4} + 4 u ^{3} - 8 u ^{2} - 4 u + 3}{u - 1.21230 \dots} = 3 u^{3} + 7.63690 \dots u^{2} + 1.25824 \dots u - 2.47462 \dots

3 u^{3} + 7.63690 \dots u^{2} + 1.25824 \dots u - 2.47462 \dots \approx 0

Encontrar una solución para

3 u^{3} + 7.63690 \dots u^{2} + 1.25824 \dots u - 2.47462 \dots = 0

utilizando el método de Newton-Raphson:

u \approx 0.45880 \dots

3 u^{3} + 7.63690 \dots u^{2} + 1.25824 \dots u - 2.47462 \dots = 0

Definición del método de Newton-Raphson

f (u) = 3 u^{3} + 7.63690 \dots u^{2} + 1.25824 \dots u - 2.47462 \dots

Hallar

f^{^{'}} (u) : 9 u^{2} + 15.27381 \dots u + 1.25824 \dots

\frac{d}{d u} (3 u^{3} + 7.63690 \dots u^{2} + 1.25824 \dots u - 2.47462 \dots)

Aplicar la regla de la suma/diferencia:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (3 u^{3}) + \frac{d}{d u} (7.63690 \dots u^{2}) + \frac{d}{d u} (1.25824 \dots u) - \frac{d}{d u} (2.47462 \dots)

\frac{d}{d u} (3 u^{3}) = 9 u^{2}

\frac{d}{d u} (3 u^{3})

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d u} (u^{3})

Aplicar la regla de la potencia:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 \cdot 3 u^{3 - 1}

Simplificar

= 9 u^{2}

\frac{d}{d u} (7.63690 \dots u^{2}) = 15.27381 \dots u

\frac{d}{d u} (7.63690 \dots u^{2})

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 7.63690 \dots \frac{d}{d u} (u^{2})

Aplicar la regla de la potencia:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 7.63690 \dots \cdot 2 u^{2 - 1}

Simplificar

= 15.27381 \dots u

\frac{d}{d u} (1.25824 \dots u) = 1.25824 \dots

\frac{d}{d u} (1.25824 \dots u)

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 1.25824 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

\frac{d u}{d u} = 1

= 1.25824 \dots \cdot 1

Simplificar

= 1.25824 \dots

\frac{d}{d u} (2.47462 \dots) = 0

\frac{d}{d u} (2.47462 \dots)

Derivada de una constante:

\frac{d}{d x} (a) = 0

= 0

= 9 u^{2} + 15.27381 \dots u + 1.25824 \dots - 0

Simplificar

= 9 u^{2} + 15.27381 \dots u + 1.25824 \dots

Sea

u_{0} = 2

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 1.19491 \dots : Δ u_{1} = 0.80508 \dots

f (u_{0}) = 3 \cdot 2^{3} + 7.63690 \dots \cdot 2^{2} + 1.25824 \dots \cdot 2 - 2.47462 \dots = 54.58948 \dots

f^{^{'}} (u_{0}) = 9 \cdot 2^{2} + 15.27381 \dots \cdot 2 + 1.25824 \dots = 67.80587 \dots

u_{1} = 1.19491 \dots

Δ u_{1} = ∣ 1.19491 \dots - 2 ∣ = 0.80508 \dots

Δ u_{1} = 0.80508 \dots

u_{2} = 0.72978 \dots : Δ u_{2} = 0.46512 \dots

f (u_{1}) = 3 \cdot 1.19491 \dots^{3} + 7.63690 \dots \cdot 1.19491 \dots^{2} + 1.25824 \dots \cdot 1.19491 \dots - 2.47462 \dots = 15.05138 \dots

f^{^{'}} (u_{1}) = 9 \cdot 1.19491 \dots^{2} + 15.27381 \dots \cdot 1.19491 \dots + 1.25824 \dots = 32.35955 \dots

u_{2} = 0.72978 \dots

Δ u_{2} = ∣ 0.72978 \dots - 1.19491 \dots ∣ = 0.46512 \dots

Δ u_{2} = 0.46512 \dots

u_{3} = 0.51598 \dots : Δ u_{3} = 0.21379 \dots

f (u_{2}) = 3 \cdot 0.72978 \dots^{3} + 7.63690 \dots \cdot 0.72978 \dots^{2} + 1.25824 \dots \cdot 0.72978 \dots - 2.47462 \dots = 3.67695 \dots

f^{^{'}} (u_{2}) = 9 \cdot 0.72978 \dots^{2} + 15.27381 \dots \cdot 0.72978 \dots + 1.25824 \dots = 17.19813 \dots

u_{3} = 0.51598 \dots

Δ u_{3} = ∣ 0.51598 \dots - 0.72978 \dots ∣ = 0.21379 \dots

Δ u_{3} = 0.21379 \dots

u_{4} = 0.46223 \dots : Δ u_{4} = 0.05374 \dots

f (u_{3}) = 3 \cdot 0.51598 \dots^{3} + 7.63690 \dots \cdot 0.51598 \dots^{2} + 1.25824 \dots \cdot 0.51598 \dots - 2.47462 \dots = 0.61999 \dots

f^{^{'}} (u_{3}) = 9 \cdot 0.51598 \dots^{2} + 15.27381 \dots \cdot 0.51598 \dots + 1.25824 \dots = 11.53548 \dots

u_{4} = 0.46223 \dots

Δ u_{4} = ∣ 0.46223 \dots - 0.51598 \dots ∣ = 0.05374 \dots

Δ u_{4} = 0.05374 \dots

u_{5} = 0.45882 \dots : Δ u_{5} = 0.00341 \dots

f (u_{4}) = 3 \cdot 0.46223 \dots^{3} + 7.63690 \dots \cdot 0.46223 \dots^{2} + 1.25824 \dots \cdot 0.46223 \dots - 2.47462 \dots = 0.03500 \dots

f^{^{'}} (u_{4}) = 9 \cdot 0.46223 \dots^{2} + 15.27381 \dots \cdot 0.46223 \dots + 1.25824 \dots = 10.24137 \dots

u_{5} = 0.45882 \dots

Δ u_{5} = ∣ 0.45882 \dots - 0.46223 \dots ∣ = 0.00341 \dots

Δ u_{5} = 0.00341 \dots

u_{6} = 0.45880 \dots : Δ u_{6} = 0.00001 \dots

f (u_{5}) = 3 \cdot 0.45882 \dots^{3} + 7.63690 \dots \cdot 0.45882 \dots^{2} + 1.25824 \dots \cdot 0.45882 \dots - 2.47462 \dots = 0.00013 \dots

f^{^{'}} (u_{5}) = 9 \cdot 0.45882 \dots^{2} + 15.27381 \dots \cdot 0.45882 \dots + 1.25824 \dots = 10.16082 \dots

u_{6} = 0.45880 \dots

Δ u_{6} = ∣ 0.45880 \dots - 0.45882 \dots ∣ = 0.00001 \dots

Δ u_{6} = 0.00001 \dots

u_{7} = 0.45880 \dots : Δ u_{7} = 2.12808 E - 10

f (u_{6}) = 3 \cdot 0.45880 \dots^{3} + 7.63690 \dots \cdot 0.45880 \dots^{2} + 1.25824 \dots \cdot 0.45880 \dots - 2.47462 \dots = 2.16224 E - 9

f^{^{'}} (u_{6}) = 9 \cdot 0.45880 \dots^{2} + 15.27381 \dots \cdot 0.45880 \dots + 1.25824 \dots = 10.16050 \dots

u_{7} = 0.45880 \dots

Δ u_{7} = ∣ 0.45880 \dots - 0.45880 \dots ∣ = 2.12808 E - 10

Δ u_{7} = 2.12808 E - 10

u \approx 0.45880 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{3 u ^{3} + 7.63690 \dots u ^{2} + 1.25824 \dots u - 2.47462 \dots}{u - 0.45880 \dots} = 3 u^{2} + 9.01332 \dots u + 5.39361 \dots

3 u^{2} + 9.01332 \dots u + 5.39361 \dots \approx 0

Encontrar una solución para

3 u^{2} + 9.01332 \dots u + 5.39361 \dots = 0

utilizando el método de Newton-Raphson:

u \approx - 0.82487 \dots

3 u^{2} + 9.01332 \dots u + 5.39361 \dots = 0

Definición del método de Newton-Raphson

f (u) = 3 u^{2} + 9.01332 \dots u + 5.39361 \dots

Hallar

f^{^{'}} (u) : 6 u + 9.01332 \dots

\frac{d}{d u} (3 u^{2} + 9.01332 \dots u + 5.39361 \dots)

Aplicar la regla de la suma/diferencia:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (3 u^{2}) + \frac{d}{d u} (9.01332 \dots u) + \frac{d}{d u} (5.39361 \dots)

\frac{d}{d u} (3 u^{2}) = 6 u

\frac{d}{d u} (3 u^{2})

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d u} (u^{2})

Aplicar la regla de la potencia:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 \cdot 2 u^{2 - 1}

Simplificar

= 6 u

\frac{d}{d u} (9.01332 \dots u) = 9.01332 \dots

\frac{d}{d u} (9.01332 \dots u)

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 9.01332 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

\frac{d u}{d u} = 1

= 9.01332 \dots \cdot 1

Simplificar

= 9.01332 \dots

\frac{d}{d u} (5.39361 \dots) = 0

\frac{d}{d u} (5.39361 \dots)

Derivada de una constante:

\frac{d}{d x} (a) = 0

= 0

= 6 u + 9.01332 \dots + 0

Simplificar

= 6 u + 9.01332 \dots

Sea

u_{0} = - 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 0.79434 \dots : Δ u_{1} = 0.20565 \dots

f (u_{0}) = 3 (- 1)^{2} + 9.01332 \dots (- 1) + 5.39361 \dots = - 0.61970 \dots

f^{^{'}} (u_{0}) = 6 (- 1) + 9.01332 \dots = 3.01332 \dots

u_{1} = - 0.79434 \dots

Δ u_{1} = ∣ - 0.79434 \dots - (- 1) ∣ = 0.20565 \dots

Δ u_{1} = 0.20565 \dots

u_{2} = - 0.82421 \dots : Δ u_{2} = 0.02987 \dots

f (u_{1}) = 3 (- 0.79434 \dots)^{2} + 9.01332 \dots (- 0.79434 \dots) + 5.39361 \dots = 0.12688 \dots

f^{^{'}} (u_{1}) = 6 (- 0.79434 \dots) + 9.01332 \dots = 4.24726 \dots

u_{2} = - 0.82421 \dots

Δ u_{2} = ∣ - 0.82421 \dots - (- 0.79434 \dots) ∣ = 0.02987 \dots

Δ u_{2} = 0.02987 \dots

u_{3} = - 0.82487 \dots : Δ u_{3} = 0.00065 \dots

f (u_{2}) = 3 (- 0.82421 \dots)^{2} + 9.01332 \dots (- 0.82421 \dots) + 5.39361 \dots = 0.00267 \dots

f^{^{'}} (u_{2}) = 6 (- 0.82421 \dots) + 9.01332 \dots = 4.06801 \dots

u_{3} = - 0.82487 \dots

Δ u_{3} = ∣ - 0.82487 \dots - (- 0.82421 \dots) ∣ = 0.00065 \dots

Δ u_{3} = 0.00065 \dots

u_{4} = - 0.82487 \dots : Δ u_{4} = 3.19753 E - 7

f (u_{3}) = 3 (- 0.82487 \dots)^{2} + 9.01332 \dots (- 0.82487 \dots) + 5.39361 \dots = 1.2995 E - 6

f^{^{'}} (u_{3}) = 6 (- 0.82487 \dots) + 9.01332 \dots = 4.06407 \dots

u_{4} = - 0.82487 \dots

Δ u_{4} = ∣ - 0.82487 \dots - (- 0.82487 \dots) ∣ = 3.19753 E - 7

Δ u_{4} = 3.19753 E - 7

u \approx - 0.82487 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{3 u ^{2} + 9.01332 \dots u + 5.39361 \dots}{u + 0.82487 \dots} = 3 u + 6.53869 \dots

3 u + 6.53869 \dots \approx 0

u \approx - 2.17956 \dots

Las soluciones son

u \approx 1.21230 \dots, u \approx 0.45880 \dots, u \approx - 0.82487 \dots, u \approx - 2.17956 \dots

u \approx 1.21230 \dots, u \approx 0.45880 \dots, u \approx - 0.82487 \dots, u \approx - 2.17956 \dots

Verificar las soluciones

Encontrar los puntos no definidos (singularidades):

u = 0

Tomar el(los) denominador(es) de

6 (\frac{u + u ^{- 1}}{2})^{2} + 4 \frac{u - u ^{- 1}}{2}

y comparar con cero

u = 0

Los siguientes puntos no están definidos

u = 0

Combinar los puntos no definidos con las soluciones:

u \approx 1.21230 \dots, u \approx 0.45880 \dots, u \approx - 0.82487 \dots, u \approx - 2.17956 \dots

u \approx 1.21230 \dots, u \approx 0.45880 \dots, u \approx - 0.82487 \dots, u \approx - 2.17956 \dots

Sustituir hacia atrás la

u = e^{x},

resolver para

x

Resolver

e^{x} = 1.21230 \dots : x = ln (1.21230 \dots)

e^{x} = 1.21230 \dots

Aplicar las leyes de los exponentes

e^{x} = 1.21230 \dots

f (x) = g (x)

, entonces

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (1.21230 \dots)

Aplicar las propiedades de los logaritmos:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1.21230 \dots)

x = ln (1.21230 \dots)

Resolver

e^{x} = 0.45880 \dots : x = ln (0.45880 \dots)

e^{x} = 0.45880 \dots

Aplicar las leyes de los exponentes

e^{x} = 0.45880 \dots

f (x) = g (x)

, entonces

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (0.45880 \dots)

Aplicar las propiedades de los logaritmos:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (0.45880 \dots)

x = ln (0.45880 \dots)

Resolver

e^{x} = - 0.82487 \dots :

Sin solución para

x \in R

e^{x} = - 0.82487 \dots

a^{f (x)}

no puede ser cero o negativo para

x \in R

Sin soluci \overset{o}{ˊ} n para x \in R

Resolver

e^{x} = - 2.17956 \dots :

Sin solución para

x \in R

e^{x} = - 2.17956 \dots

a^{f (x)}

no puede ser cero o negativo para

x \in R

Sin soluci \overset{o}{ˊ} n para x \in R

x = ln (1.21230 \dots), x = ln (0.45880 \dots)

x = ln (1.21230 \dots), x = ln (0.45880 \dots)

Gráfica

Ejemplos populares

5sqrt(3)tan(x)+3=8sqrt(3)tan(x)

53 tan (x) + 3 = 83 tan (x)

cos^2(x)=-0.5

cos^{2} (x) = - 0.5

cos(x)=(1.5)/(4.272)

cos (x) = \frac{1.5}{4.272}

pi/(12)=arcsin(x/2)

\frac{π}{12} = arcsin (\frac{x}{2})

6sin^2(x)=0

6 sin^{2} (x) = 0