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Trazado el gráfico de la función/ ((sin2x)/(1+cosx))+(1/(ln(1+sinx)))

Gráfico de la función y = ((sin2x)/(1+cosx))+(1/(ln(1+sinx)))

v

Gráfico:

interior superior

Puntos de intersección:

mostrar?

Definida a trozos:

Solución

Ha introducido [src] 
        sin(2*x)           1       
f(x) = ---------- + ---------------
       1 + cos(x)   log(1 + sin(x))
f(x)=1log⁡(sin⁡(x)+1)+sin⁡(2x)cos⁡(x)+1f{\left(x \right)} = \frac{1}{\log{\left(\sin{\left(x \right)} + 1 \right)}} + \frac{\sin{\left(2 x \right)}}{\cos{\left(x \right)} + 1} 
f = 1/log(sin(x) + 1) + sin(2*x)/(cos(x) + 1)
Gráfico de la función
02468-8-6-4-2-1010-10001000
Dominio de definición de la función
Puntos en los que la función no está definida exactamente:
x1=0x_{1} = 0
x2=3.14159265358979x_{2} = 3.14159265358979
Puntos de cruce con el eje de coordenadas X
El gráfico de la función cruce el eje X con f = 0
o sea hay que resolver la ecuación:
1log⁡(sin⁡(x)+1)+sin⁡(2x)cos⁡(x)+1=0\frac{1}{\log{\left(\sin{\left(x \right)} + 1 \right)}} + \frac{\sin{\left(2 x \right)}}{\cos{\left(x \right)} + 1} = 0
Resolvermos esta ecuación
Puntos de cruce con el eje X:

Solución numérica
x1=58.6124295473767x_{1} = 58.6124295473767
x2=77.4619854689155x_{2} = 77.4619854689155
x3=−54.4849059818559x_{3} = -54.4849059818559
x4=54.8972985721627x_{4} = 54.8972985721627
x5=67.4636691865219x_{5} = 67.4636691865219
x6=39.7628736258379x_{6} = 39.7628736258379
x7=−1.65136919245356x_{7} = -1.65136919245356
x8=27.1965030114788x_{8} = 27.1965030114788
x9=−20.5009251139923x_{9} = -20.5009251139923
x10=8.34694708994001x_{10} = 8.34694708994001
x11=64.8956148545563x_{11} = 64.8956148545563
x12=−83.3327781857882x_{12} = -83.3327781857882
x13=−16.7857941387783x_{13} = -16.7857941387783
x14=−33.0672957283515x_{14} = -33.0672957283515
x15=−51.9168516498902x_{15} = -51.9168516498902
x16=61.1804838793423x_{16} = 61.1804838793423
x17=−79.6176472105742x_{17} = -79.6176472105742
x18=20.9133177042992x_{18} = 20.9133177042992
x19=92.5964104152402x_{19} = 92.5964104152402
x20=−26.7841104211719x_{20} = -26.7841104211719
x21=−10.5026088315988x_{21} = -10.5026088315988
x22=29.7645573434444x_{22} = 29.7645573434444
x23=−14.2177398068127x_{23} = -14.2177398068127
x24=98.8795957224198x_{24} = 98.8795957224198
x25=−23.0689794459579x_{25} = -23.0689794459579
x26=46.0460589330175x_{26} = 46.0460589330175
x27=83.7451707760951x_{27} = 83.7451707760951
x28=48.6141132649831x_{28} = 48.6141132649831
x29=52.3292442401971x_{29} = 52.3292442401971
x30=42.3309279578035x_{30} = 42.3309279578035
x31=−89.6159634929678x_{31} = -89.6159634929678
x32=14.6301323971196x_{32} = 14.6301323971196
x33=−39.3504810355311x_{33} = -39.3504810355311
x34=−58.2000369570698x_{34} = -58.2000369570698
x35=36.047742650624x_{35} = 36.047742650624
x36=−70.766407571429x_{36} = -70.766407571429
x37=71.1788001617359x_{37} = 71.1788001617359
x38=23.4813720362648x_{38} = 23.4813720362648
x39=−35.6353500603171x_{39} = -35.6353500603171
x40=17.1981867290852x_{40} = 17.1981867290852
x41=−92.1840178249334x_{41} = -92.1840178249334
x42=73.7468544937015x_{42} = 73.7468544937015
x43=−77.0495928786086x_{43} = -77.0495928786086
x44=−41.9185353674967x_{44} = -41.9185353674967
x45=−73.3344619033946x_{45} = -73.3344619033946
x46=86.3132251080606x_{46} = 86.3132251080606
x47=−85.9008325177538x_{47} = -85.9008325177538
x48=−60.7680912890354x_{48} = -60.7680912890354
x49=2.06376178276042x_{49} = 2.06376178276042
x50=80.0300398008811x_{50} = 80.0300398008811
x51=−7.93455449963314x_{51} = -7.93455449963314
x52=−45.6336663427107x_{52} = -45.6336663427107
x53=−4.21942352441916x_{53} = -4.21942352441916
x54=−64.4832222642494x_{54} = -64.4832222642494
x55=4.63181611472603x_{55} = 4.63181611472603
x56=33.4796883186584x_{56} = 33.4796883186584
x57=10.9150014219056x_{57} = 10.9150014219056
x58=−48.2017206746763x_{58} = -48.2017206746763
x59=−29.3521647531375x_{59} = -29.3521647531375
x60=96.3115413904542x_{60} = 96.3115413904542
x61=90.0283560832746x_{61} = 90.0283560832746
x62=−98.467203132113x_{62} = -98.467203132113
x63=−95.8991488001474x_{63} = -95.8991488001474
x64=−67.051276596215x_{64} = -67.051276596215
Asíntotas verticales
Hay:
x1=0x_{1} = 0
x2=3.14159265358979x_{2} = 3.14159265358979
Asíntotas horizontales
Hallemos las asíntotas horizontales mediante los límites de esta función con x->+oo y x->-oo
True

Tomamos como el límite
es decir,
ecuación de la asíntota horizontal a la izquierda:
y=lim⁡x→−∞(1log⁡(sin⁡(x)+1)+sin⁡(2x)cos⁡(x)+1)y = \lim_{x \to -\infty}\left(\frac{1}{\log{\left(\sin{\left(x \right)} + 1 \right)}} + \frac{\sin{\left(2 x \right)}}{\cos{\left(x \right)} + 1}\right)
True

Tomamos como el límite
es decir,
ecuación de la asíntota horizontal a la derecha:
y=lim⁡x→∞(1log⁡(sin⁡(x)+1)+sin⁡(2x)cos⁡(x)+1)y = \lim_{x \to \infty}\left(\frac{1}{\log{\left(\sin{\left(x \right)} + 1 \right)}} + \frac{\sin{\left(2 x \right)}}{\cos{\left(x \right)} + 1}\right)
Asíntotas inclinadas
Se puede hallar la asíntota inclinada calculando el límite de la función sin(2*x)/(1 + cos(x)) + 1/log(1 + sin(x)), dividida por x con x->+oo y x ->-oo
True

Tomamos como el límite
es decir,
ecuación de la asíntota inclinada a la izquierda:
y=xlim⁡x→−∞(1log⁡(sin⁡(x)+1)+sin⁡(2x)cos⁡(x)+1x)y = x \lim_{x \to -\infty}\left(\frac{\frac{1}{\log{\left(\sin{\left(x \right)} + 1 \right)}} + \frac{\sin{\left(2 x \right)}}{\cos{\left(x \right)} + 1}}{x}\right)
True

Tomamos como el límite
es decir,
ecuación de la asíntota inclinada a la derecha:
y=xlim⁡x→∞(1log⁡(sin⁡(x)+1)+sin⁡(2x)cos⁡(x)+1x)y = x \lim_{x \to \infty}\left(\frac{\frac{1}{\log{\left(\sin{\left(x \right)} + 1 \right)}} + \frac{\sin{\left(2 x \right)}}{\cos{\left(x \right)} + 1}}{x}\right)
Paridad e imparidad de la función
Comprobemos si la función es par o impar mediante las relaciones f = f(-x) и f = -f(-x).
Pues, comprobamos:
1log⁡(sin⁡(x)+1)+sin⁡(2x)cos⁡(x)+1=1log⁡(1−sin⁡(x))−sin⁡(2x)cos⁡(x)+1\frac{1}{\log{\left(\sin{\left(x \right)} + 1 \right)}} + \frac{\sin{\left(2 x \right)}}{\cos{\left(x \right)} + 1} = \frac{1}{\log{\left(1 - \sin{\left(x \right)} \right)}} - \frac{\sin{\left(2 x \right)}}{\cos{\left(x \right)} + 1}
- No
1log⁡(sin⁡(x)+1)+sin⁡(2x)cos⁡(x)+1=−1log⁡(1−sin⁡(x))+sin⁡(2x)cos⁡(x)+1\frac{1}{\log{\left(\sin{\left(x \right)} + 1 \right)}} + \frac{\sin{\left(2 x \right)}}{\cos{\left(x \right)} + 1} = - \frac{1}{\log{\left(1 - \sin{\left(x \right)} \right)}} + \frac{\sin{\left(2 x \right)}}{\cos{\left(x \right)} + 1}
- No
es decir, función
no es
par ni impar
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