English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometría >

2cosh(2x)=5sinh(x)+3

Solución

2 cosh (2 x) = 5 sinh (x) + 3

Solución

x = ln (0.83987 \dots), x = ln (3.16657 \dots)

Grados

x = - 9.99831 \dots^{\circ}, x = 66.04208 \dots^{\circ}

Pasos de solución

2 cosh (2 x) = 5 sinh (x) + 3

Re-escribir usando identidades trigonométricas

2 cosh (2 x) = 5 sinh (x) + 3

Utilizar la identidad hiperbólica:

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

2 cosh (2 x) = 5 \cdot \frac{e ^{x} - e ^{- x}}{2} + 3

Utilizar la identidad hiperbólica:

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

2 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 5 \cdot \frac{e ^{x} - e ^{- x}}{2} + 3

2 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 5 \cdot \frac{e ^{x} - e ^{- x}}{2} + 3

2 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 5 \cdot \frac{e ^{x} - e ^{- x}}{2} + 3 : x = ln (0.83987 \dots), x = ln (3.16657 \dots)

2 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 5 \cdot \frac{e ^{x} - e ^{- x}}{2} + 3

Aplicar las leyes de los exponentes

2 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 5 \cdot \frac{e ^{x} - e ^{- x}}{2} + 3

Aplicar las leyes de los exponentes:

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

e^{2 x} = (e^{x})^{2}, e^{- 2 x} = (e^{x})^{- 2}, e^{- x} = (e^{x})^{- 1}

2 \cdot \frac{( e ^{x} ) ^{2} + ( e ^{x} ) ^{- 2}}{2} = 5 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} + 3

2 \cdot \frac{( e ^{x} ) ^{2} + ( e ^{x} ) ^{- 2}}{2} = 5 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} + 3

Re escribir la ecuación con

e^{x} = u

2 \cdot \frac{( u ) ^{2} + ( u ) ^{- 2}}{2} = 5 \cdot \frac{u - ( u ) ^{- 1}}{2} + 3

Resolver

2 \cdot \frac{u ^{2} + u ^{- 2}}{2} = 5 \cdot \frac{u - u ^{- 1}}{2} + 3 : u \approx - 0.31579 \dots, u \approx - 1.19065 \dots, u \approx 0.83987 \dots, u \approx 3.16657 \dots

2 \cdot \frac{u ^{2} + u ^{- 2}}{2} = 5 \cdot \frac{u - u ^{- 1}}{2} + 3

Simplificar

u^{2} + \frac{1}{u ^{2}} = \frac{5 ( u ^{2} - 1 )}{2 u} + 3

Multiplicar por el mínimo común múltiplo

u^{2} + \frac{1}{u ^{2}} = \frac{5 ( u ^{2} - 1 )}{2 u} + 3

Encontrar el mínimo común múltiplo de

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

u^{2}, 2 u

Mínimo común múltiplo (MCM)

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

u^{2}

2 u

= 2 u^{2}

Multiplicar por el mínimo común múltiplo=

2 u^{2}

u^{2} \cdot 2 u^{2} + \frac{1}{u ^{2}} \cdot 2 u^{2} = \frac{5 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2} + 3 \cdot 2 u^{2}

Simplificar

u^{2} \cdot 2 u^{2} + \frac{1}{u ^{2}} \cdot 2 u^{2} = \frac{5 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2} + 3 \cdot 2 u^{2}

Simplificar

u^{2} \cdot 2 u^{2} : 2 u^{4}

u^{2} \cdot 2 u^{2}

Aplicar las leyes de los exponentes:

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

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

= 2 u^{2 + 2}

Sumar:

2 + 2 = 4

= 2 u^{4}

Simplificar

\frac{1}{u ^{2}} \cdot 2 u^{2} : 2

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

Multiplicar fracciones:

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

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

Eliminar los terminos comunes:

u^{2}

= 1 \cdot 2

Multiplicar los numeros:

1 \cdot 2 = 2

= 2

Simplificar

\frac{5 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2} : 5 u (u^{2} - 1)

\frac{5 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2}

Multiplicar fracciones:

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

= \frac{5 ( u ^{2} - 1 ) \cdot 2 u ^{2}}{2 u}

Eliminar los terminos comunes:

2

= \frac{5 ( u ^{2} - 1 ) u ^{2}}{u}

Eliminar los terminos comunes:

u

= 5 u (u^{2} - 1)

Simplificar

3 \cdot 2 u^{2} : 6 u^{2}

3 \cdot 2 u^{2}

Multiplicar los numeros:

3 \cdot 2 = 6

= 6 u^{2}

2 u^{4} + 2 = 5 u (u^{2} - 1) + 6 u^{2}

2 u^{4} + 2 = 5 u (u^{2} - 1) + 6 u^{2}

2 u^{4} + 2 = 5 u (u^{2} - 1) + 6 u^{2}

Resolver

2 u^{4} + 2 = 5 u (u^{2} - 1) + 6 u^{2} : u \approx - 0.31579 \dots, u \approx - 1.19065 \dots, u \approx 0.83987 \dots, u \approx 3.16657 \dots

2 u^{4} + 2 = 5 u (u^{2} - 1) + 6 u^{2}

Desarrollar

5 u (u^{2} - 1) + 6 u^{2} : 5 u^{3} - 5 u + 6 u^{2}

5 u (u^{2} - 1) + 6 u^{2}

Expandir

5 u (u^{2} - 1) : 5 u^{3} - 5 u

5 u (u^{2} - 1)

Poner los parentesis utilizando:

a (b - c) = ab - a c

a = 5 u, b = u^{2}, c = 1

= 5 u u^{2} - 5 u \cdot 1

= 5 u^{2} u - 5 \cdot 1 \cdot u

Simplificar

5 u^{2} u - 5 \cdot 1 \cdot u : 5 u^{3} - 5 u

5 u^{2} u - 5 \cdot 1 \cdot u

5 u^{2} u = 5 u^{3}

5 u^{2} u

Aplicar las leyes de los exponentes:

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

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

= 5 u^{2 + 1}

Sumar:

2 + 1 = 3

= 5 u^{3}

5 \cdot 1 \cdot u = 5 u

5 \cdot 1 \cdot u

Multiplicar los numeros:

5 \cdot 1 = 5

= 5 u

= 5 u^{3} - 5 u

= 5 u^{3} - 5 u

= 5 u^{3} - 5 u + 6 u^{2}

2 u^{4} + 2 = 5 u^{3} - 5 u + 6 u^{2}

Desplace

6 u^{2}

a la izquierda

2 u^{4} + 2 = 5 u^{3} - 5 u + 6 u^{2}

Restar

6 u^{2}

de ambos lados

2 u^{4} + 2 - 6 u^{2} = 5 u^{3} - 5 u + 6 u^{2} - 6 u^{2}

Simplificar

2 u^{4} + 2 - 6 u^{2} = 5 u^{3} - 5 u

2 u^{4} + 2 - 6 u^{2} = 5 u^{3} - 5 u

Desplace

5 u

a la izquierda

2 u^{4} + 2 - 6 u^{2} = 5 u^{3} - 5 u

Sumar

5 u

a ambos lados

2 u^{4} + 2 - 6 u^{2} + 5 u = 5 u^{3} - 5 u + 5 u

Simplificar

2 u^{4} + 2 - 6 u^{2} + 5 u = 5 u^{3}

2 u^{4} + 2 - 6 u^{2} + 5 u = 5 u^{3}

Desplace

5 u^{3}

a la izquierda

2 u^{4} + 2 - 6 u^{2} + 5 u = 5 u^{3}

Restar

5 u^{3}

de ambos lados

2 u^{4} + 2 - 6 u^{2} + 5 u - 5 u^{3} = 5 u^{3} - 5 u^{3}

Simplificar

2 u^{4} + 2 - 6 u^{2} + 5 u - 5 u^{3} = 0

2 u^{4} + 2 - 6 u^{2} + 5 u - 5 u^{3} = 0

Escribir en la forma binómica

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

2 u^{4} - 5 u^{3} - 6 u^{2} + 5 u + 2 = 0

Encontrar una solución para

2 u^{4} - 5 u^{3} - 6 u^{2} + 5 u + 2 = 0

utilizando el método de Newton-Raphson:

u \approx - 0.31579 \dots

2 u^{4} - 5 u^{3} - 6 u^{2} + 5 u + 2 = 0

Definición del método de Newton-Raphson

f (u) = 2 u^{4} - 5 u^{3} - 6 u^{2} + 5 u + 2

Hallar

f^{^{'}} (u) : 8 u^{3} - 15 u^{2} - 12 u + 5

\frac{d}{d u} (2 u^{4} - 5 u^{3} - 6 u^{2} + 5 u + 2)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (2 u^{4}) - \frac{d}{d u} (5 u^{3}) - \frac{d}{d u} (6 u^{2}) + \frac{d}{d u} (5 u) + \frac{d}{d u} (2)

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

\frac{d}{d u} (2 u^{4})

Sacar la constante:

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

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

Aplicar la regla de la potencia:

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

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

Simplificar

= 8 u^{3}

\frac{d}{d u} (5 u^{3}) = 15 u^{2}

\frac{d}{d u} (5 u^{3})

Sacar la constante:

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

= 5 \frac{d}{d u} (u^{3})

Aplicar la regla de la potencia:

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

= 5 \cdot 3 u^{3 - 1}

Simplificar

= 15 u^{2}

\frac{d}{d u} (6 u^{2}) = 12 u

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

Sacar la constante:

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

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

Aplicar la regla de la potencia:

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

= 6 \cdot 2 u^{2 - 1}

Simplificar

= 12 u

\frac{d}{d u} (5 u) = 5

\frac{d}{d u} (5 u)

Sacar la constante:

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

= 5 \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 5 \cdot 1

Simplificar

= 5

\frac{d}{d u} (2) = 0

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

Derivada de una constante:

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

= 0

= 8 u^{3} - 15 u^{2} - 12 u + 5 + 0

Simplificar

= 8 u^{3} - 15 u^{2} - 12 u + 5

Sea

u_{0} = 0

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 0.4 : Δ u_{1} = 0.4

f (u_{0}) = 2 \cdot 0^{4} - 5 \cdot 0^{3} - 6 \cdot 0^{2} + 5 \cdot 0 + 2 = 2

f^{^{'}} (u_{0}) = 8 \cdot 0^{3} - 15 \cdot 0^{2} - 12 \cdot 0 + 5 = 5

u_{1} = - 0.4

Δ u_{1} = ∣ - 0.4 - 0 ∣ = 0.4

Δ u_{1} = 0.4

u_{2} = - 0.31451 \dots : Δ u_{2} = 0.08548 \dots

f (u_{1}) = 2 (- 0.4)^{4} - 5 (- 0.4)^{3} - 6 (- 0.4)^{2} + 5 (- 0.4) + 2 = - 0.5888

f^{^{'}} (u_{1}) = 8 (- 0.4)^{3} - 15 (- 0.4)^{2} - 12 (- 0.4) + 5 = 6.888

u_{2} = - 0.31451 \dots

Δ u_{2} = ∣ - 0.31451 \dots - (- 0.4) ∣ = 0.08548 \dots

Δ u_{2} = 0.08548 \dots

u_{3} = - 0.31579 \dots : Δ u_{3} = 0.00128 \dots

f (u_{2}) = 2 (- 0.31451 \dots)^{4} - 5 (- 0.31451 \dots)^{3} - 6 (- 0.31451 \dots)^{2} + 5 (- 0.31451 \dots) + 2 = 0.00901 \dots

f^{^{'}} (u_{2}) = 8 (- 0.31451 \dots)^{3} - 15 (- 0.31451 \dots)^{2} - 12 (- 0.31451 \dots) + 5 = 7.04149 \dots

u_{3} = - 0.31579 \dots

Δ u_{3} = ∣ - 0.31579 \dots - (- 0.31451 \dots) ∣ = 0.00128 \dots

Δ u_{3} = 0.00128 \dots

u_{4} = - 0.31579 \dots : Δ u_{4} = 1.99105 E - 8

f (u_{3}) = 2 (- 0.31579 \dots)^{4} - 5 (- 0.31579 \dots)^{3} - 6 (- 0.31579 \dots)^{2} + 5 (- 0.31579 \dots) + 2 = - 1.40204 E - 7

f^{^{'}} (u_{3}) = 8 (- 0.31579 \dots)^{3} - 15 (- 0.31579 \dots)^{2} - 12 (- 0.31579 \dots) + 5 = 7.04169 \dots

u_{4} = - 0.31579 \dots

Δ u_{4} = ∣ - 0.31579 \dots - (- 0.31579 \dots) ∣ = 1.99105 E - 8

Δ u_{4} = 1.99105 E - 8

u \approx - 0.31579 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{2 u ^{4} - 5 u ^{3} - 6 u ^{2} + 5 u + 2}{u + 0.31579 \dots} = 2 u^{3} - 5.63159 \dots u^{2} - 4.22155 \dots u + 6.33315 \dots

2 u^{3} - 5.63159 \dots u^{2} - 4.22155 \dots u + 6.33315 \dots \approx 0

Encontrar una solución para

2 u^{3} - 5.63159 \dots u^{2} - 4.22155 \dots u + 6.33315 \dots = 0

utilizando el método de Newton-Raphson:

u \approx - 1.19065 \dots

2 u^{3} - 5.63159 \dots u^{2} - 4.22155 \dots u + 6.33315 \dots = 0

Definición del método de Newton-Raphson

f (u) = 2 u^{3} - 5.63159 \dots u^{2} - 4.22155 \dots u + 6.33315 \dots

Hallar

f^{^{'}} (u) : 6 u^{2} - 11.26319 \dots u - 4.22155 \dots

\frac{d}{d u} (2 u^{3} - 5.63159 \dots u^{2} - 4.22155 \dots u + 6.33315 \dots)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (2 u^{3}) - \frac{d}{d u} (5.63159 \dots u^{2}) - \frac{d}{d u} (4.22155 \dots u) + \frac{d}{d u} (6.33315 \dots)

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

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

Sacar la constante:

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

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

Aplicar la regla de la potencia:

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

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

Simplificar

= 6 u^{2}

\frac{d}{d u} (5.63159 \dots u^{2}) = 11.26319 \dots u

\frac{d}{d u} (5.63159 \dots u^{2})

Sacar la constante:

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

= 5.63159 \dots \frac{d}{d u} (u^{2})

Aplicar la regla de la potencia:

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

= 5.63159 \dots \cdot 2 u^{2 - 1}

Simplificar

= 11.26319 \dots u

\frac{d}{d u} (4.22155 \dots u) = 4.22155 \dots

\frac{d}{d u} (4.22155 \dots u)

Sacar la constante:

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

= 4.22155 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 4.22155 \dots \cdot 1

Simplificar

= 4.22155 \dots

\frac{d}{d u} (6.33315 \dots) = 0

\frac{d}{d u} (6.33315 \dots)

Derivada de una constante:

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

= 0

= 6 u^{2} - 11.26319 \dots u - 4.22155 \dots + 0

Simplificar

= 6 u^{2} - 11.26319 \dots u - 4.22155 \dots

Sea

u_{0} = 2

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 1.14284 \dots : Δ u_{1} = 3.14284 \dots

f (u_{0}) = 2 \cdot 2^{3} - 5.63159 \dots \cdot 2^{2} - 4.22155 \dots \cdot 2 + 6.33315 \dots = - 8.63633 \dots

f^{^{'}} (u_{0}) = 6 \cdot 2^{2} - 11.26319 \dots \cdot 2 - 4.22155 \dots = - 2.74793 \dots

u_{1} = - 1.14284 \dots

Δ u_{1} = ∣ - 1.14284 \dots - 2 ∣ = 3.14284 \dots

Δ u_{1} = 3.14284 \dots

u_{2} = - 1.19239 \dots : Δ u_{2} = 0.04955 \dots

f (u_{1}) = 2 (- 1.14284 \dots)^{3} - 5.63159 \dots (- 1.14284 \dots)^{2} - 4.22155 \dots (- 1.14284 \dots) + 6.33315 \dots = 0.81707 \dots

f^{^{'}} (u_{1}) = 6 (- 1.14284 \dots)^{2} - 11.26319 \dots (- 1.14284 \dots) - 4.22155 \dots = 16.48700 \dots

u_{2} = - 1.19239 \dots

Δ u_{2} = ∣ - 1.19239 \dots - (- 1.14284 \dots) ∣ = 0.04955 \dots

Δ u_{2} = 0.04955 \dots

u_{3} = - 1.19065 \dots : Δ u_{3} = 0.00174 \dots

f (u_{2}) = 2 (- 1.19239 \dots)^{3} - 5.63159 \dots (- 1.19239 \dots)^{2} - 4.22155 \dots (- 1.19239 \dots) + 6.33315 \dots = - 0.03091 \dots

f^{^{'}} (u_{2}) = 6 (- 1.19239 \dots)^{2} - 11.26319 \dots (- 1.19239 \dots) - 4.22155 \dots = 17.73958 \dots

u_{3} = - 1.19065 \dots

Δ u_{3} = ∣ - 1.19065 \dots - (- 1.19239 \dots) ∣ = 0.00174 \dots

Δ u_{3} = 0.00174 \dots

u_{4} = - 1.19065 \dots : Δ u_{4} = 2.19412 E - 6

f (u_{3}) = 2 (- 1.19065 \dots)^{3} - 5.63159 \dots (- 1.19065 \dots)^{2} - 4.22155 \dots (- 1.19065 \dots) + 6.33315 \dots = - 0.00003 \dots

f^{^{'}} (u_{3}) = 6 (- 1.19065 \dots)^{2} - 11.26319 \dots (- 1.19065 \dots) - 4.22155 \dots = 17.69503 \dots

u_{4} = - 1.19065 \dots

Δ u_{4} = ∣ - 1.19065 \dots - (- 1.19065 \dots) ∣ = 2.19412 E - 6

Δ u_{4} = 2.19412 E - 6

u_{5} = - 1.19065 \dots : Δ u_{5} = 3.47577 E - 12

f (u_{4}) = 2 (- 1.19065 \dots)^{3} - 5.63159 \dots (- 1.19065 \dots)^{2} - 4.22155 \dots (- 1.19065 \dots) + 6.33315 \dots = - 6.15037 E - 11

f^{^{'}} (u_{4}) = 6 (- 1.19065 \dots)^{2} - 11.26319 \dots (- 1.19065 \dots) - 4.22155 \dots = 17.69497 \dots

u_{5} = - 1.19065 \dots

Δ u_{5} = ∣ - 1.19065 \dots - (- 1.19065 \dots) ∣ = 3.47577 E - 12

Δ u_{5} = 3.47577 E - 12

u \approx - 1.19065 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{2 u ^{3} - 5.63159 \dots u ^{2} - 4.22155 \dots u + 6.33315 \dots}{u + 1.19065 \dots} = 2 u^{2} - 8.01290 \dots u + 5.31905 \dots

2 u^{2} - 8.01290 \dots u + 5.31905 \dots \approx 0

Encontrar una solución para

2 u^{2} - 8.01290 \dots u + 5.31905 \dots = 0

utilizando el método de Newton-Raphson:

u \approx 0.83987 \dots

2 u^{2} - 8.01290 \dots u + 5.31905 \dots = 0

Definición del método de Newton-Raphson

f (u) = 2 u^{2} - 8.01290 \dots u + 5.31905 \dots

Hallar

f^{^{'}} (u) : 4 u - 8.01290 \dots

\frac{d}{d u} (2 u^{2} - 8.01290 \dots u + 5.31905 \dots)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (2 u^{2}) - \frac{d}{d u} (8.01290 \dots u) + \frac{d}{d u} (5.31905 \dots)

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

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

Sacar la constante:

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

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

Aplicar la regla de la potencia:

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

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

Simplificar

= 4 u

\frac{d}{d u} (8.01290 \dots u) = 8.01290 \dots

\frac{d}{d u} (8.01290 \dots u)

Sacar la constante:

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

= 8.01290 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 8.01290 \dots \cdot 1

Simplificar

= 8.01290 \dots

\frac{d}{d u} (5.31905 \dots) = 0

\frac{d}{d u} (5.31905 \dots)

Derivada de una constante:

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

= 0

= 4 u - 8.01290 \dots + 0

Simplificar

= 4 u - 8.01290 \dots

Sea

u_{0} = 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 0.82709 \dots : Δ u_{1} = 0.17290 \dots

f (u_{0}) = 2 \cdot 1^{2} - 8.01290 \dots \cdot 1 + 5.31905 \dots = - 0.69385 \dots

f^{^{'}} (u_{0}) = 4 \cdot 1 - 8.01290 \dots = - 4.01290 \dots

u_{1} = 0.82709 \dots

Δ u_{1} = ∣ 0.82709 \dots - 1 ∣ = 0.17290 \dots

Δ u_{1} = 0.17290 \dots

u_{2} = 0.83980 \dots : Δ u_{2} = 0.01270 \dots

f (u_{1}) = 2 \cdot 0.82709 \dots^{2} - 8.01290 \dots \cdot 0.82709 \dots + 5.31905 \dots = 0.05979 \dots

f^{^{'}} (u_{1}) = 4 \cdot 0.82709 \dots - 8.01290 \dots = - 4.70452 \dots

u_{2} = 0.83980 \dots

Δ u_{2} = ∣ 0.83980 \dots - 0.82709 \dots ∣ = 0.01270 \dots

Δ u_{2} = 0.01270 \dots

u_{3} = 0.83987 \dots : Δ u_{3} = 0.00006 \dots

f (u_{2}) = 2 \cdot 0.83980 \dots^{2} - 8.01290 \dots \cdot 0.83980 \dots + 5.31905 \dots = 0.00032 \dots

f^{^{'}} (u_{2}) = 4 \cdot 0.83980 \dots - 8.01290 \dots = - 4.65368 \dots

u_{3} = 0.83987 \dots

Δ u_{3} = ∣ 0.83987 \dots - 0.83980 \dots ∣ = 0.00006 \dots

Δ u_{3} = 0.00006 \dots

u_{4} = 0.83987 \dots : Δ u_{4} = 2.0713 E - 9

f (u_{3}) = 2 \cdot 0.83987 \dots^{2} - 8.01290 \dots \cdot 0.83987 \dots + 5.31905 \dots = 9.63859 E - 9

f^{^{'}} (u_{3}) = 4 \cdot 0.83987 \dots - 8.01290 \dots = - 4.65341 \dots

u_{4} = 0.83987 \dots

Δ u_{4} = ∣ 0.83987 \dots - 0.83987 \dots ∣ = 2.0713 E - 9

Δ u_{4} = 2.0713 E - 9

u \approx 0.83987 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{2 u ^{2} - 8.01290 \dots u + 5.31905 \dots}{u - 0.83987 \dots} = 2 u - 6.33315 \dots

2 u - 6.33315 \dots \approx 0

u \approx 3.16657 \dots

Las soluciones son

u \approx - 0.31579 \dots, u \approx - 1.19065 \dots, u \approx 0.83987 \dots, u \approx 3.16657 \dots

u \approx - 0.31579 \dots, u \approx - 1.19065 \dots, u \approx 0.83987 \dots, u \approx 3.16657 \dots

Verificar las soluciones

Encontrar los puntos no definidos (singularidades):

u = 0

Tomar el(los) denominador(es) de

2 \frac{u ^{2} + u ^{- 2}}{2}

y comparar con cero

Resolver

u^{2} = 0 : u = 0

u^{2} = 0

Aplicar la regla

x^{n} = 0 \Rightarrow x = 0

u = 0

Tomar el(los) denominador(es) de

5 \frac{u - u ^{- 1}}{2} + 3

y comparar con cero

u = 0

Los siguientes puntos no están definidos

u = 0

Combinar los puntos no definidos con las soluciones:

u \approx - 0.31579 \dots, u \approx - 1.19065 \dots, u \approx 0.83987 \dots, u \approx 3.16657 \dots

u \approx - 0.31579 \dots, u \approx - 1.19065 \dots, u \approx 0.83987 \dots, u \approx 3.16657 \dots

Sustituir hacia atrás la

u = e^{x},

resolver para

x

Resolver

e^{x} = - 0.31579 \dots :

Sin solución para

x \in R

e^{x} = - 0.31579 \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} = - 1.19065 \dots :

Sin solución para

x \in R

e^{x} = - 1.19065 \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} = 0.83987 \dots : x = ln (0.83987 \dots)

e^{x} = 0.83987 \dots

Aplicar las leyes de los exponentes

e^{x} = 0.83987 \dots

f (x) = g (x)

, entonces

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

ln (e^{x}) = ln (0.83987 \dots)

Aplicar las propiedades de los logaritmos:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (0.83987 \dots)

x = ln (0.83987 \dots)

Resolver

e^{x} = 3.16657 \dots : x = ln (3.16657 \dots)

e^{x} = 3.16657 \dots

Aplicar las leyes de los exponentes

e^{x} = 3.16657 \dots

f (x) = g (x)

, entonces

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

ln (e^{x}) = ln (3.16657 \dots)

Aplicar las propiedades de los logaritmos:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (3.16657 \dots)

x = ln (3.16657 \dots)

x = ln (0.83987 \dots), x = ln (3.16657 \dots)

x = ln (0.83987 \dots), x = ln (3.16657 \dots)

Gráfica

Ejemplos populares

tan(x)=(6.9)/(6.6)

sin(x)-cos(x)=(sqrt(2))/2

2cos(x)=7-3/(cos(x))

csc(x)=-9/7 ,tan(x)>0

sin(x)= 12/37