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Popular Trigonometría >

tan^{22}(x)=sec^2(x)-1

Solución

tan^{22} (x) = sec^{2} (x) - 1

Solución

x = πn, x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Grados

x = 0^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Pasos de solución

tan^{22} (x) = sec^{2} (x) - 1

Restar

sec^{2} (x) - 1

de ambos lados

tan^{22} (x) - sec^{2} (x) + 1 = 0

Re-escribir usando identidades trigonométricas

1 - sec^{2} (x) + tan^{22} (x)

Utilizar la identidad pitagórica:

sec^{2} (x) = tan^{2} (x) + 1

sec^{2} (x) - 1 = tan^{2} (x)

= tan^{22} (x) - tan^{2} (x)

tan^{22} (x) - tan^{2} (x) = 0

Usando el método de sustitución

tan^{22} (x) - tan^{2} (x) = 0

Sea:

tan (x) = u

u^{22} - u^{2} = 0

u^{22} - u^{2} = 0 : u = 0, u = 1, u = - 1

u^{22} - u^{2} = 0

Re-escribir la ecuación con

v = u^{2}

v^{11} = u^{22}

v^{11} - v = 0

Resolver

v^{11} - v = 0 : v = 0, v = - 1, v = 1

v^{11} - v = 0

Factorizar

v^{11} - v : v (v + 1) (v^{4} - v^{3} + v^{2} - v + 1) (v - 1) (v^{4} + v^{3} + v^{2} + v + 1)

v^{11} - v

Factorizar el termino común

v : v (v^{10} - 1)

v^{11} - v

Aplicar las leyes de los exponentes:

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

v^{11} = v^{10} v

= v^{10} v - v

Factorizar el termino común

v

= v (v^{10} - 1)

= v (v^{10} - 1)

Factorizar

v^{10} - 1 : (v + 1) (v^{4} - v^{3} + v^{2} - v + 1) (v - 1) (v^{4} + v^{3} + v^{2} + v + 1)

v^{10} - 1

Reescribir

v^{10} - 1

como

(v^{5})^{2} - 1^{2}

v^{10} - 1

Reescribir

1

como

1^{2}

= v^{10} - 1^{2}

Aplicar las leyes de los exponentes:

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

v^{10} = (v^{5})^{2}

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

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

Aplicar la siguiente regla para binomios al cuadrado:

x^{2} - y^{2} = (x + y) (x - y)

(v^{5})^{2} - 1^{2} = (v^{5} + 1) (v^{5} - 1)

= (v^{5} + 1) (v^{5} - 1)

Factorizar

v^{5} + 1 : (v + 1) (v^{4} - v^{3} + v^{2} - v + 1)

v^{5} + 1

Reescribir

1

como

1^{5}

= v^{5} + 1^{5}

Aplicar la regla de factorización:

x^{n} + y^{n} = (x + y) (x^{n - 1} - x^{n - 2} y + \dots - x y^{n - 2} + y^{n - 1})

n is odd

v^{5} + 1^{5} = (v + 1) (v^{4} - v^{3} + v^{2} - v + 1)

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

= (v + 1) (v^{4} - v^{3} + v^{2} - v + 1) (v^{5} - 1)

Factorizar

v^{5} - 1 : (v - 1) (v^{4} + v^{3} + v^{2} + v + 1)

v^{5} - 1

Reescribir

1

como

1^{5}

= v^{5} - 1^{5}

Aplicar la regla de factorización:

x^{n} - y^{n} = (x - y) (x^{n - 1} + x^{n - 2} y + \dots + x y^{n - 2} y^{n - 1})

v^{5} - 1^{5} = (v - 1) (v^{4} + v^{3} + v^{2} + v + 1)

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

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

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

v (v + 1) (v^{4} - v^{3} + v^{2} - v + 1) (v - 1) (v^{4} + v^{3} + v^{2} + v + 1) = 0

Usando la propiedad del factor cero:

ab = 0

entonces

a = 0

b = 0

v = 0 or v + 1 = 0 or v^{4} - v^{3} + v^{2} - v + 1 = 0 or v - 1 = 0 or v^{4} + v^{3} + v^{2} + v + 1 = 0

Resolver

v + 1 = 0 : v = - 1

v + 1 = 0

Desplace

1

a la derecha

v + 1 = 0

Restar

1

de ambos lados

v + 1 - 1 = 0 - 1

Simplificar

v = - 1

v = - 1

Resolver

v^{4} - v^{3} + v^{2} - v + 1 = 0 :

Sin solución para

v \in R

v^{4} - v^{3} + v^{2} - v + 1 = 0

Encontrar una solución para

v^{4} - v^{3} + v^{2} - v + 1 = 0

utilizando el método de Newton-Raphson:

Sin solución para

v \in R

v^{4} - v^{3} + v^{2} - v + 1 = 0

Definición del método de Newton-Raphson

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

Hallar

f^{^{'}} (v) : 4 v^{3} - 3 v^{2} + 2 v - 1

\frac{d}{d v} (v^{4} - v^{3} + v^{2} - v + 1)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d v} (v^{4}) - \frac{d}{d v} (v^{3}) + \frac{d}{d v} (v^{2}) - \frac{d v}{d v} + \frac{d}{d v} (1)

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

\frac{d}{d v} (v^{4})

Aplicar la regla de la potencia:

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

= 4 v^{4 - 1}

Simplificar

= 4 v^{3}

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

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

Aplicar la regla de la potencia:

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

= 3 v^{3 - 1}

Simplificar

= 3 v^{2}

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

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

Aplicar la regla de la potencia:

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

= 2 v^{2 - 1}

Simplificar

= 2 v

\frac{d v}{d v} = 1

\frac{d v}{d v}

Aplicar la regla de derivación:

\frac{d v}{d v} = 1

= 1

\frac{d}{d v} (1) = 0

\frac{d}{d v} (1)

Derivada de una constante:

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

= 0

= 4 v^{3} - 3 v^{2} + 2 v - 1 + 0

Simplificar

= 4 v^{3} - 3 v^{2} + 2 v - 1

Sea

v_{0} = 1

Calcular

v_{n + 1}

hasta que

Δ v_{n + 1} < 0.000001

v_{1} = 0.5 : Δ v_{1} = 0.5

f (v_{0}) = 1^{4} - 1^{3} + 1^{2} - 1 + 1 = 1

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

v_{1} = 0.5

Δ v_{1} = ∣ 0.5 - 1 ∣ = 0.5

Δ v_{1} = 0.5

v_{2} = 3.25 : Δ v_{2} = 2.75

f (v_{1}) = 0. 5^{4} - 0. 5^{3} + 0. 5^{2} - 0.5 + 1 = 0.6875

f^{^{'}} (v_{1}) = 4 \cdot 0. 5^{3} - 3 \cdot 0. 5^{2} + 2 \cdot 0.5 - 1 = - 0.25

v_{2} = 3.25

Δ v_{2} = ∣ 3.25 - 0.5 ∣ = 2.75

Δ v_{2} = 2.75

v_{3} = 2.48013 \dots : Δ v_{3} = 0.76986 \dots

f (v_{2}) = 3.2 5^{4} - 3.2 5^{3} + 3.2 5^{2} - 3.25 + 1 = 85.55078 \dots

f^{^{'}} (v_{2}) = 4 \cdot 3.2 5^{3} - 3 \cdot 3.2 5^{2} + 2 \cdot 3.25 - 1 = 111.125

v_{3} = 2.48013 \dots

Δ v_{3} = ∣ 2.48013 \dots - 3.25 ∣ = 0.76986 \dots

Δ v_{3} = 0.76986 \dots

v_{4} = 1.89445 \dots : Δ v_{4} = 0.58568 \dots

f (v_{3}) = 2.48013 \dots^{4} - 2.48013 \dots^{3} + 2.48013 \dots^{2} - 2.48013 \dots + 1 = 27.25130 \dots

f^{^{'}} (v_{3}) = 4 \cdot 2.48013 \dots^{3} - 3 \cdot 2.48013 \dots^{2} + 2 \cdot 2.48013 \dots - 1 = 46.52924 \dots

v_{4} = 1.89445 \dots

Δ v_{4} = ∣ 1.89445 \dots - 2.48013 \dots ∣ = 0.58568 \dots

Δ v_{4} = 0.58568 \dots

v_{5} = 1.43781 \dots : Δ v_{5} = 0.45664 \dots

f (v_{4}) = 1.89445 \dots^{4} - 1.89445 \dots^{3} + 1.89445 \dots^{2} - 1.89445 \dots + 1 = 8.77607 \dots

f^{^{'}} (v_{4}) = 4 \cdot 1.89445 \dots^{3} - 3 \cdot 1.89445 \dots^{2} + 2 \cdot 1.89445 \dots - 1 = 19.21862 \dots

v_{5} = 1.43781 \dots

Δ v_{5} = ∣ 1.43781 \dots - 1.89445 \dots ∣ = 0.45664 \dots

Δ v_{5} = 0.45664 \dots

v_{6} = 1.05030 \dots : Δ v_{6} = 0.38750 \dots

f (v_{5}) = 1.43781 \dots^{4} - 1.43781 \dots^{3} + 1.43781 \dots^{2} - 1.43781 \dots + 1 = 2.93085 \dots

f^{^{'}} (v_{5}) = 4 \cdot 1.43781 \dots^{3} - 3 \cdot 1.43781 \dots^{2} + 2 \cdot 1.43781 \dots - 1 = 7.56332 \dots

v_{6} = 1.05030 \dots

Δ v_{6} = ∣ 1.05030 \dots - 1.43781 \dots ∣ = 0.38750 \dots

Δ v_{6} = 0.38750 \dots

v_{7} = 0.59224 \dots : Δ v_{7} = 0.45805 \dots

f (v_{6}) = 1.05030 \dots^{4} - 1.05030 \dots^{3} + 1.05030 \dots^{2} - 1.05030 \dots + 1 = 1.11112 \dots

f^{^{'}} (v_{6}) = 4 \cdot 1.05030 \dots^{3} - 3 \cdot 1.05030 \dots^{2} + 2 \cdot 1.05030 \dots - 1 = 2.42572 \dots

v_{7} = 0.59224 \dots

Δ v_{7} = ∣ 0.59224 \dots - 1.05030 \dots ∣ = 0.45805 \dots

Δ v_{7} = 0.45805 \dots

v_{8} = 18.88435 \dots : Δ v_{8} = 18.29210 \dots

f (v_{7}) = 0.59224 \dots^{4} - 0.59224 \dots^{3} + 0.59224 \dots^{2} - 0.59224 \dots + 1 = 0.67380 \dots

f^{^{'}} (v_{7}) = 4 \cdot 0.59224 \dots^{3} - 3 \cdot 0.59224 \dots^{2} + 2 \cdot 0.59224 \dots - 1 = - 0.03683 \dots

v_{8} = 18.88435 \dots

Δ v_{8} = ∣ 18.88435 \dots - 0.59224 \dots ∣ = 18.29210 \dots

Δ v_{8} = 18.29210 \dots

v_{9} = 14.22188 \dots : Δ v_{9} = 4.66247 \dots

f (v_{8}) = 18.88435 \dots^{4} - 18.88435 \dots^{3} + 18.88435 \dots^{2} - 18.88435 \dots + 1 = 120781.11894 \dots

f^{^{'}} (v_{8}) = 4 \cdot 18.88435 \dots^{3} - 3 \cdot 18.88435 \dots^{2} + 2 \cdot 18.88435 \dots - 1 = 25904.96293 \dots

v_{9} = 14.22188 \dots

Δ v_{9} = ∣ 14.22188 \dots - 18.88435 \dots ∣ = 4.66247 \dots

Δ v_{9} = 4.66247 \dots

v_{10} = 10.72385 \dots : Δ v_{10} = 3.49802 \dots

f (v_{9}) = 14.22188 \dots^{4} - 14.22188 \dots^{3} + 14.22188 \dots^{2} - 14.22188 \dots + 1 = 38222.36483 \dots

f^{^{'}} (v_{9}) = 4 \cdot 14.22188 \dots^{3} - 3 \cdot 14.22188 \dots^{2} + 2 \cdot 14.22188 \dots - 1 = 10926.83534 \dots

v_{10} = 10.72385 \dots

Δ v_{10} = ∣ 10.72385 \dots - 14.22188 \dots ∣ = 3.49802 \dots

Δ v_{10} = 3.49802 \dots

v_{11} = 8.09884 \dots : Δ v_{11} = 2.62501 \dots

f (v_{10}) = 10.72385 \dots^{4} - 10.72385 \dots^{3} + 10.72385 \dots^{2} - 10.72385 \dots + 1 = 12097.26043 \dots

f^{^{'}} (v_{10}) = 4 \cdot 10.72385 \dots^{3} - 3 \cdot 10.72385 \dots^{2} + 2 \cdot 10.72385 \dots - 1 = 4608.46147 \dots

v_{11} = 8.09884 \dots

Δ v_{11} = ∣ 8.09884 \dots - 10.72385 \dots ∣ = 2.62501 \dots

Δ v_{11} = 2.62501 \dots

v_{12} = 6.12820 \dots : Δ v_{12} = 1.97063 \dots

f (v_{11}) = 8.09884 \dots^{4} - 8.09884 \dots^{3} + 8.09884 \dots^{2} - 8.09884 \dots + 1 = 3829.49182 \dots

f^{^{'}} (v_{11}) = 4 \cdot 8.09884 \dots^{3} - 3 \cdot 8.09884 \dots^{2} + 2 \cdot 8.09884 \dots - 1 = 1943.27695 \dots

v_{12} = 6.12820 \dots

Δ v_{12} = ∣ 6.12820 \dots - 8.09884 \dots ∣ = 1.97063 \dots

Δ v_{12} = 1.97063 \dots

v_{13} = 4.64785 \dots : Δ v_{13} = 1.48034 \dots

f (v_{12}) = 6.12820 \dots^{4} - 6.12820 \dots^{3} + 6.12820 \dots^{2} - 6.12820 \dots + 1 = 1212.65418 \dots

f^{^{'}} (v_{12}) = 4 \cdot 6.12820 \dots^{3} - 3 \cdot 6.12820 \dots^{2} + 2 \cdot 6.12820 \dots - 1 = 819.16882 \dots

v_{13} = 4.64785 \dots

Δ v_{13} = ∣ 4.64785 \dots - 6.12820 \dots ∣ = 1.48034 \dots

Δ v_{13} = 1.48034 \dots

v_{14} = 3.53453 \dots : Δ v_{14} = 1.11332 \dots

f (v_{13}) = 4.64785 \dots^{4} - 4.64785 \dots^{3} + 4.64785 \dots^{2} - 4.64785 \dots + 1 = 384.22115 \dots

f^{^{'}} (v_{13}) = 4 \cdot 4.64785 \dots^{3} - 3 \cdot 4.64785 \dots^{2} + 2 \cdot 4.64785 \dots - 1 = 345.11129 \dots

v_{14} = 3.53453 \dots

Δ v_{14} = ∣ 3.53453 \dots - 4.64785 \dots ∣ = 1.11332 \dots

Δ v_{14} = 1.11332 \dots

v_{15} = 2.69527 \dots : Δ v_{15} = 0.83926 \dots

f (v_{14}) = 3.53453 \dots^{4} - 3.53453 \dots^{3} + 3.53453 \dots^{2} - 3.53453 \dots + 1 = 121.87505 \dots

f^{^{'}} (v_{14}) = 4 \cdot 3.53453 \dots^{3} - 3 \cdot 3.53453 \dots^{2} + 2 \cdot 3.53453 \dots - 1 = 145.21705 \dots

v_{15} = 2.69527 \dots

Δ v_{15} = ∣ 2.69527 \dots - 3.53453 \dots ∣ = 0.83926 \dots

Δ v_{15} = 0.83926 \dots

v_{16} = 2.05895 \dots : Δ v_{16} = 0.63632 \dots

f (v_{15}) = 2.69527 \dots^{4} - 2.69527 \dots^{3} + 2.69527 \dots^{2} - 2.69527 \dots + 1 = 38.76232 \dots

f^{^{'}} (v_{15}) = 4 \cdot 2.69527 \dots^{3} - 3 \cdot 2.69527 \dots^{2} + 2 \cdot 2.69527 \dots - 1 = 60.91625 \dots

v_{16} = 2.05895 \dots

Δ v_{16} = ∣ 2.05895 \dots - 2.69527 \dots ∣ = 0.63632 \dots

Δ v_{16} = 0.63632 \dots

v_{17} = 1.56818 \dots : Δ v_{17} = 0.49077 \dots

f (v_{16}) = 2.05895 \dots^{4} - 2.05895 \dots^{3} + 2.05895 \dots^{2} - 2.05895 \dots + 1 = 12.42335 \dots

f^{^{'}} (v_{16}) = 4 \cdot 2.05895 \dots^{3} - 3 \cdot 2.05895 \dots^{2} + 2 \cdot 2.05895 \dots - 1 = 25.31395 \dots

v_{17} = 1.56818 \dots

Δ v_{17} = ∣ 1.56818 \dots - 2.05895 \dots ∣ = 0.49077 \dots

Δ v_{17} = 0.49077 \dots

v_{18} = 1.16736 \dots : Δ v_{18} = 0.40081 \dots

f (v_{17}) = 1.56818 \dots^{4} - 1.56818 \dots^{3} + 1.56818 \dots^{2} - 1.56818 \dots + 1 = 4.08216 \dots

f^{^{'}} (v_{17}) = 4 \cdot 1.56818 \dots^{3} - 3 \cdot 1.56818 \dots^{2} + 2 \cdot 1.56818 \dots - 1 = 10.18459 \dots

v_{18} = 1.16736 \dots

Δ v_{18} = ∣ 1.16736 \dots - 1.56818 \dots ∣ = 0.40081 \dots

Δ v_{18} = 0.40081 \dots

v_{19} = 0.76245 \dots : Δ v_{19} = 0.40490 \dots

f (v_{18}) = 1.16736 \dots^{4} - 1.16736 \dots^{3} + 1.16736 \dots^{2} - 1.16736 \dots + 1 = 1.46161 \dots

f^{^{'}} (v_{18}) = 4 \cdot 1.16736 \dots^{3} - 3 \cdot 1.16736 \dots^{2} + 2 \cdot 1.16736 \dots - 1 = 3.60974 \dots

v_{19} = 0.76245 \dots

Δ v_{19} = ∣ 0.76245 \dots - 1.16736 \dots ∣ = 0.40490 \dots

Δ v_{19} = 0.40490 \dots

No se puede encontrar solución

La solución es

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

Resolver

v - 1 = 0 : v = 1

v - 1 = 0

Desplace

1

a la derecha

v - 1 = 0

Sumar

1

a ambos lados

v - 1 + 1 = 0 + 1

Simplificar

v = 1

v = 1

Resolver

v^{4} + v^{3} + v^{2} + v + 1 = 0 :

Sin solución para

v \in R

v^{4} + v^{3} + v^{2} + v + 1 = 0

Encontrar una solución para

v^{4} + v^{3} + v^{2} + v + 1 = 0

utilizando el método de Newton-Raphson:

Sin solución para

v \in R

v^{4} + v^{3} + v^{2} + v + 1 = 0

Definición del método de Newton-Raphson

f (v) = v^{4} + v^{3} + v^{2} + v + 1

Hallar

f^{^{'}} (v) : 4 v^{3} + 3 v^{2} + 2 v + 1

\frac{d}{d v} (v^{4} + v^{3} + v^{2} + v + 1)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d v} (v^{4}) + \frac{d}{d v} (v^{3}) + \frac{d}{d v} (v^{2}) + \frac{d v}{d v} + \frac{d}{d v} (1)

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

\frac{d}{d v} (v^{4})

Aplicar la regla de la potencia:

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

= 4 v^{4 - 1}

Simplificar

= 4 v^{3}

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

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

Aplicar la regla de la potencia:

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

= 3 v^{3 - 1}

Simplificar

= 3 v^{2}

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

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

Aplicar la regla de la potencia:

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

= 2 v^{2 - 1}

Simplificar

= 2 v

\frac{d v}{d v} = 1

\frac{d v}{d v}

Aplicar la regla de derivación:

\frac{d v}{d v} = 1

= 1

\frac{d}{d v} (1) = 0

\frac{d}{d v} (1)

Derivada de una constante:

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

= 0

= 4 v^{3} + 3 v^{2} + 2 v + 1 + 0

Simplificar

= 4 v^{3} + 3 v^{2} + 2 v + 1

Sea

v_{0} = - 1

Calcular

v_{n + 1}

hasta que

Δ v_{n + 1} < 0.000001

v_{1} = - 0.5 : Δ v_{1} = 0.5

f (v_{0}) = (- 1)^{4} + (- 1)^{3} + (- 1)^{2} + (- 1) + 1 = 1

f^{^{'}} (v_{0}) = 4 (- 1)^{3} + 3 (- 1)^{2} + 2 (- 1) + 1 = - 2

v_{1} = - 0.5

Δ v_{1} = ∣ - 0.5 - (- 1) ∣ = 0.5

Δ v_{1} = 0.5

v_{2} = - 3.25 : Δ v_{2} = 2.75

f (v_{1}) = (- 0.5)^{4} + (- 0.5)^{3} + (- 0.5)^{2} + (- 0.5) + 1 = 0.6875

f^{^{'}} (v_{1}) = 4 (- 0.5)^{3} + 3 (- 0.5)^{2} + 2 (- 0.5) + 1 = 0.25

v_{2} = - 3.25

Δ v_{2} = ∣ - 3.25 - (- 0.5) ∣ = 2.75

Δ v_{2} = 2.75

v_{3} = - 2.48013 \dots : Δ v_{3} = 0.76986 \dots

f (v_{2}) = (- 3.25)^{4} + (- 3.25)^{3} + (- 3.25)^{2} + (- 3.25) + 1 = 85.55078 \dots

f^{^{'}} (v_{2}) = 4 (- 3.25)^{3} + 3 (- 3.25)^{2} + 2 (- 3.25) + 1 = - 111.125

v_{3} = - 2.48013 \dots

Δ v_{3} = ∣ - 2.48013 \dots - (- 3.25) ∣ = 0.76986 \dots

Δ v_{3} = 0.76986 \dots

v_{4} = - 1.89445 \dots : Δ v_{4} = 0.58568 \dots

f (v_{3}) = (- 2.48013 \dots)^{4} + (- 2.48013 \dots)^{3} + (- 2.48013 \dots)^{2} + (- 2.48013 \dots) + 1 = 27.25130 \dots

f^{^{'}} (v_{3}) = 4 (- 2.48013 \dots)^{3} + 3 (- 2.48013 \dots)^{2} + 2 (- 2.48013 \dots) + 1 = - 46.52924 \dots

v_{4} = - 1.89445 \dots

Δ v_{4} = ∣ - 1.89445 \dots - (- 2.48013 \dots) ∣ = 0.58568 \dots

Δ v_{4} = 0.58568 \dots

v_{5} = - 1.43781 \dots : Δ v_{5} = 0.45664 \dots

f (v_{4}) = (- 1.89445 \dots)^{4} + (- 1.89445 \dots)^{3} + (- 1.89445 \dots)^{2} + (- 1.89445 \dots) + 1 = 8.77607 \dots

f^{^{'}} (v_{4}) = 4 (- 1.89445 \dots)^{3} + 3 (- 1.89445 \dots)^{2} + 2 (- 1.89445 \dots) + 1 = - 19.21862 \dots

v_{5} = - 1.43781 \dots

Δ v_{5} = ∣ - 1.43781 \dots - (- 1.89445 \dots) ∣ = 0.45664 \dots

Δ v_{5} = 0.45664 \dots

v_{6} = - 1.05030 \dots : Δ v_{6} = 0.38750 \dots

f (v_{5}) = (- 1.43781 \dots)^{4} + (- 1.43781 \dots)^{3} + (- 1.43781 \dots)^{2} + (- 1.43781 \dots) + 1 = 2.93085 \dots

f^{^{'}} (v_{5}) = 4 (- 1.43781 \dots)^{3} + 3 (- 1.43781 \dots)^{2} + 2 (- 1.43781 \dots) + 1 = - 7.56332 \dots

v_{6} = - 1.05030 \dots

Δ v_{6} = ∣ - 1.05030 \dots - (- 1.43781 \dots) ∣ = 0.38750 \dots

Δ v_{6} = 0.38750 \dots

v_{7} = - 0.59224 \dots : Δ v_{7} = 0.45805 \dots

f (v_{6}) = (- 1.05030 \dots)^{4} + (- 1.05030 \dots)^{3} + (- 1.05030 \dots)^{2} + (- 1.05030 \dots) + 1 = 1.11112 \dots

f^{^{'}} (v_{6}) = 4 (- 1.05030 \dots)^{3} + 3 (- 1.05030 \dots)^{2} + 2 (- 1.05030 \dots) + 1 = - 2.42572 \dots

v_{7} = - 0.59224 \dots

Δ v_{7} = ∣ - 0.59224 \dots - (- 1.05030 \dots) ∣ = 0.45805 \dots

Δ v_{7} = 0.45805 \dots

v_{8} = - 18.88435 \dots : Δ v_{8} = 18.29210 \dots

f (v_{7}) = (- 0.59224 \dots)^{4} + (- 0.59224 \dots)^{3} + (- 0.59224 \dots)^{2} + (- 0.59224 \dots) + 1 = 0.67380 \dots

f^{^{'}} (v_{7}) = 4 (- 0.59224 \dots)^{3} + 3 (- 0.59224 \dots)^{2} + 2 (- 0.59224 \dots) + 1 = 0.03683 \dots

v_{8} = - 18.88435 \dots

Δ v_{8} = ∣ - 18.88435 \dots - (- 0.59224 \dots) ∣ = 18.29210 \dots

Δ v_{8} = 18.29210 \dots

v_{9} = - 14.22188 \dots : Δ v_{9} = 4.66247 \dots

f (v_{8}) = (- 18.88435 \dots)^{4} + (- 18.88435 \dots)^{3} + (- 18.88435 \dots)^{2} + (- 18.88435 \dots) + 1 = 120781.11894 \dots

f^{^{'}} (v_{8}) = 4 (- 18.88435 \dots)^{3} + 3 (- 18.88435 \dots)^{2} + 2 (- 18.88435 \dots) + 1 = - 25904.96293 \dots

v_{9} = - 14.22188 \dots

Δ v_{9} = ∣ - 14.22188 \dots - (- 18.88435 \dots) ∣ = 4.66247 \dots

Δ v_{9} = 4.66247 \dots

v_{10} = - 10.72385 \dots : Δ v_{10} = 3.49802 \dots

f (v_{9}) = (- 14.22188 \dots)^{4} + (- 14.22188 \dots)^{3} + (- 14.22188 \dots)^{2} + (- 14.22188 \dots) + 1 = 38222.36483 \dots

f^{^{'}} (v_{9}) = 4 (- 14.22188 \dots)^{3} + 3 (- 14.22188 \dots)^{2} + 2 (- 14.22188 \dots) + 1 = - 10926.83534 \dots

v_{10} = - 10.72385 \dots

Δ v_{10} = ∣ - 10.72385 \dots - (- 14.22188 \dots) ∣ = 3.49802 \dots

Δ v_{10} = 3.49802 \dots

v_{11} = - 8.09884 \dots : Δ v_{11} = 2.62501 \dots

f (v_{10}) = (- 10.72385 \dots)^{4} + (- 10.72385 \dots)^{3} + (- 10.72385 \dots)^{2} + (- 10.72385 \dots) + 1 = 12097.26043 \dots

f^{^{'}} (v_{10}) = 4 (- 10.72385 \dots)^{3} + 3 (- 10.72385 \dots)^{2} + 2 (- 10.72385 \dots) + 1 = - 4608.46147 \dots

v_{11} = - 8.09884 \dots

Δ v_{11} = ∣ - 8.09884 \dots - (- 10.72385 \dots) ∣ = 2.62501 \dots

Δ v_{11} = 2.62501 \dots

v_{12} = - 6.12820 \dots : Δ v_{12} = 1.97063 \dots

f (v_{11}) = (- 8.09884 \dots)^{4} + (- 8.09884 \dots)^{3} + (- 8.09884 \dots)^{2} + (- 8.09884 \dots) + 1 = 3829.49182 \dots

f^{^{'}} (v_{11}) = 4 (- 8.09884 \dots)^{3} + 3 (- 8.09884 \dots)^{2} + 2 (- 8.09884 \dots) + 1 = - 1943.27695 \dots

v_{12} = - 6.12820 \dots

Δ v_{12} = ∣ - 6.12820 \dots - (- 8.09884 \dots) ∣ = 1.97063 \dots

Δ v_{12} = 1.97063 \dots

v_{13} = - 4.64785 \dots : Δ v_{13} = 1.48034 \dots

f (v_{12}) = (- 6.12820 \dots)^{4} + (- 6.12820 \dots)^{3} + (- 6.12820 \dots)^{2} + (- 6.12820 \dots) + 1 = 1212.65418 \dots

f^{^{'}} (v_{12}) = 4 (- 6.12820 \dots)^{3} + 3 (- 6.12820 \dots)^{2} + 2 (- 6.12820 \dots) + 1 = - 819.16882 \dots

v_{13} = - 4.64785 \dots

Δ v_{13} = ∣ - 4.64785 \dots - (- 6.12820 \dots) ∣ = 1.48034 \dots

Δ v_{13} = 1.48034 \dots

v_{14} = - 3.53453 \dots : Δ v_{14} = 1.11332 \dots

f (v_{13}) = (- 4.64785 \dots)^{4} + (- 4.64785 \dots)^{3} + (- 4.64785 \dots)^{2} + (- 4.64785 \dots) + 1 = 384.22115 \dots

f^{^{'}} (v_{13}) = 4 (- 4.64785 \dots)^{3} + 3 (- 4.64785 \dots)^{2} + 2 (- 4.64785 \dots) + 1 = - 345.11129 \dots

v_{14} = - 3.53453 \dots

Δ v_{14} = ∣ - 3.53453 \dots - (- 4.64785 \dots) ∣ = 1.11332 \dots

Δ v_{14} = 1.11332 \dots

v_{15} = - 2.69527 \dots : Δ v_{15} = 0.83926 \dots

f (v_{14}) = (- 3.53453 \dots)^{4} + (- 3.53453 \dots)^{3} + (- 3.53453 \dots)^{2} + (- 3.53453 \dots) + 1 = 121.87505 \dots

f^{^{'}} (v_{14}) = 4 (- 3.53453 \dots)^{3} + 3 (- 3.53453 \dots)^{2} + 2 (- 3.53453 \dots) + 1 = - 145.21705 \dots

v_{15} = - 2.69527 \dots

Δ v_{15} = ∣ - 2.69527 \dots - (- 3.53453 \dots) ∣ = 0.83926 \dots

Δ v_{15} = 0.83926 \dots

v_{16} = - 2.05895 \dots : Δ v_{16} = 0.63632 \dots

f (v_{15}) = (- 2.69527 \dots)^{4} + (- 2.69527 \dots)^{3} + (- 2.69527 \dots)^{2} + (- 2.69527 \dots) + 1 = 38.76232 \dots

f^{^{'}} (v_{15}) = 4 (- 2.69527 \dots)^{3} + 3 (- 2.69527 \dots)^{2} + 2 (- 2.69527 \dots) + 1 = - 60.91625 \dots

v_{16} = - 2.05895 \dots

Δ v_{16} = ∣ - 2.05895 \dots - (- 2.69527 \dots) ∣ = 0.63632 \dots

Δ v_{16} = 0.63632 \dots

v_{17} = - 1.56818 \dots : Δ v_{17} = 0.49077 \dots

f (v_{16}) = (- 2.05895 \dots)^{4} + (- 2.05895 \dots)^{3} + (- 2.05895 \dots)^{2} + (- 2.05895 \dots) + 1 = 12.42335 \dots

f^{^{'}} (v_{16}) = 4 (- 2.05895 \dots)^{3} + 3 (- 2.05895 \dots)^{2} + 2 (- 2.05895 \dots) + 1 = - 25.31395 \dots

v_{17} = - 1.56818 \dots

Δ v_{17} = ∣ - 1.56818 \dots - (- 2.05895 \dots) ∣ = 0.49077 \dots

Δ v_{17} = 0.49077 \dots

v_{18} = - 1.16736 \dots : Δ v_{18} = 0.40081 \dots

f (v_{17}) = (- 1.56818 \dots)^{4} + (- 1.56818 \dots)^{3} + (- 1.56818 \dots)^{2} + (- 1.56818 \dots) + 1 = 4.08216 \dots

f^{^{'}} (v_{17}) = 4 (- 1.56818 \dots)^{3} + 3 (- 1.56818 \dots)^{2} + 2 (- 1.56818 \dots) + 1 = - 10.18459 \dots

v_{18} = - 1.16736 \dots

Δ v_{18} = ∣ - 1.16736 \dots - (- 1.56818 \dots) ∣ = 0.40081 \dots

Δ v_{18} = 0.40081 \dots

v_{19} = - 0.76245 \dots : Δ v_{19} = 0.40490 \dots

f (v_{18}) = (- 1.16736 \dots)^{4} + (- 1.16736 \dots)^{3} + (- 1.16736 \dots)^{2} + (- 1.16736 \dots) + 1 = 1.46161 \dots

f^{^{'}} (v_{18}) = 4 (- 1.16736 \dots)^{3} + 3 (- 1.16736 \dots)^{2} + 2 (- 1.16736 \dots) + 1 = - 3.60974 \dots

v_{19} = - 0.76245 \dots

Δ v_{19} = ∣ - 0.76245 \dots - (- 1.16736 \dots) ∣ = 0.40490 \dots

Δ v_{19} = 0.40490 \dots

No se puede encontrar solución

La solución es

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

Las soluciones son

v = 0, v = - 1, v = 1

v = 0, v = - 1, v = 1

Sustituir hacia atrás la

v = u^{2},

resolver para

u

Resolver

u^{2} = 0 : u = 0

u^{2} = 0

Aplicar la regla

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

u = 0

Resolver

u^{2} = - 1 :

Sin solución para

u \in R

u^{2} = - 1

x^{2}

no puede ser negativo para

x \in R

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

Resolver

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

Para

x^{2} = f (a)

las soluciones son

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Aplicar la regla

1 = 1

= 1

- 1 = - 1

- 1

Aplicar la regla

1 = 1

= - 1

u = 1, u = - 1

Las soluciones son

u = 0, u = 1, u = - 1

Sustituir en la ecuación

u = tan (x)

tan (x) = 0, tan (x) = 1, tan (x) = - 1

tan (x) = 0, tan (x) = 1, tan (x) = - 1

tan (x) = 0 : x = πn

tan (x) = 0

Soluciones generales para

tan (x) = 0

tan (x)

tabla de valores periódicos con

πn

intervalos:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + πn

x = 0 + πn

Resolver

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

Soluciones generales para

tan (x) = 1

tan (x)

tabla de valores periódicos con

πn

intervalos:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

tan (x) = - 1 : x = \frac{3 π}{4} + πn

tan (x) = - 1

Soluciones generales para

tan (x) = - 1

tan (x)

tabla de valores periódicos con

πn

intervalos:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combinar toda las soluciones

x = πn, x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Gráfica

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