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y=tg(x)-ctg(x/4)
Trazado el gráfico de la función/ y=tg(x)-ctg(x/4)

Gráfico de la función y = y=tg(x)-ctg(x/4)

v

Gráfico:

interior superior

Puntos de intersección:

mostrar?

Definida a trozos:

Solución

Ha introducido [src] 
                   /x\
f(x) = tan(x) - cot|-|
                   \4/
f(x)=tan⁡(x)−cot⁡(x4)f{\left(x \right)} = \tan{\left(x \right)} - \cot{\left(\frac{x}{4} \right)} 
f = tan(x) - cot(x/4)
Gráfico de la función
02468-8-6-4-2-1010-200200
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:
tan⁡(x)−cot⁡(x4)=0\tan{\left(x \right)} - \cot{\left(\frac{x}{4} \right)} = 0
Resolvermos esta ecuación
Puntos de cruce con el eje X:

Solución analítica
x1=−18π5x_{1} = - \frac{18 \pi}{5}
x2=−2πx_{2} = - 2 \pi
x3=2π5x_{3} = \frac{2 \pi}{5}
x4=2πx_{4} = 2 \pi
x5=−4ilog⁡(−105+58+25+58+i4+5i4)x_{5} = - 4 i \log{\left(- \frac{\sqrt{10} \sqrt{\sqrt{5} + 5}}{8} + \frac{\sqrt{2} \sqrt{\sqrt{5} + 5}}{8} + \frac{i}{4} + \frac{\sqrt{5} i}{4} \right)}
x6=−4ilog⁡(−25+58+105+58−5i4−i4)x_{6} = - 4 i \log{\left(- \frac{\sqrt{2} \sqrt{\sqrt{5} + 5}}{8} + \frac{\sqrt{10} \sqrt{\sqrt{5} + 5}}{8} - \frac{\sqrt{5} i}{4} - \frac{i}{4} \right)}
x7=−4ilog⁡(−105−516+25−516+25+516+105+516+i4+5i4)x_{7} = - 4 i \log{\left(- \frac{\sqrt{10} \sqrt{5 - \sqrt{5}}}{16} + \frac{\sqrt{2} \sqrt{5 - \sqrt{5}}}{16} + \frac{\sqrt{2} \sqrt{\sqrt{5} + 5}}{16} + \frac{\sqrt{10} \sqrt{\sqrt{5} + 5}}{16} + \frac{i}{4} + \frac{\sqrt{5} i}{4} \right)}
x8=−4ilog⁡(−105+516−105−516−25+516+25−516−i4+5i4)x_{8} = - 4 i \log{\left(- \frac{\sqrt{10} \sqrt{\sqrt{5} + 5}}{16} - \frac{\sqrt{10} \sqrt{5 - \sqrt{5}}}{16} - \frac{\sqrt{2} \sqrt{\sqrt{5} + 5}}{16} + \frac{\sqrt{2} \sqrt{5 - \sqrt{5}}}{16} - \frac{i}{4} + \frac{\sqrt{5} i}{4} \right)}
x9=−4ilog⁡(−105+516−25+516−25−516+105−516−5i4−i4)x_{9} = - 4 i \log{\left(- \frac{\sqrt{10} \sqrt{\sqrt{5} + 5}}{16} - \frac{\sqrt{2} \sqrt{\sqrt{5} + 5}}{16} - \frac{\sqrt{2} \sqrt{5 - \sqrt{5}}}{16} + \frac{\sqrt{10} \sqrt{5 - \sqrt{5}}}{16} - \frac{\sqrt{5} i}{4} - \frac{i}{4} \right)}
x10=−4ilog⁡(−25−516+25+516+105−516+105+516−5i4+i4)x_{10} = - 4 i \log{\left(- \frac{\sqrt{2} \sqrt{5 - \sqrt{5}}}{16} + \frac{\sqrt{2} \sqrt{\sqrt{5} + 5}}{16} + \frac{\sqrt{10} \sqrt{5 - \sqrt{5}}}{16} + \frac{\sqrt{10} \sqrt{\sqrt{5} + 5}}{16} - \frac{\sqrt{5} i}{4} + \frac{i}{4} \right)}
Solución numérica
x1=−21.3628300444106x_{1} = -21.3628300444106
x2=36.4424747816416x_{2} = 36.4424747816416
x3=−69.1150383789755x_{3} = -69.1150383789755
x4=28.9026524130261x_{4} = 28.9026524130261
x5=−96.7610537305656x_{5} = -96.7610537305656
x6=−41.4690230273853x_{6} = -41.4690230273853
x7=−61.5752160103599x_{7} = -61.5752160103599
x8=6.28318530717959x_{8} = 6.28318530717959
x9=−64.0884901332318x_{9} = -64.0884901332318
x10=−76.654860747591x_{10} = -76.654860747591
x11=−33.9292006587698x_{11} = -33.9292006587698
x12=−81.6814089933346x_{12} = -81.6814089933346
x13=43.9822971502571x_{13} = 43.9822971502571
x14=69.1150383789755x_{14} = 69.1150383789755
x15=−99.2743278534375x_{15} = -99.2743278534375
x16=−94.2477796076938x_{16} = -94.2477796076938
x17=76.654860747591x_{17} = 76.654860747591
x18=−38.9557489045134x_{18} = -38.9557489045134
x19=3.76991118430775x_{19} = 3.76991118430775
x20=−66.6017642561036x_{20} = -66.6017642561036
x21=−74.1415866247191x_{21} = -74.1415866247191
x22=41.4690230273853x_{22} = 41.4690230273853
x23=−71.6283125018473x_{23} = -71.6283125018473
x24=−91.734505484822x_{24} = -91.734505484822
x25=71.6283125018473x_{25} = 71.6283125018473
x26=81.6814089933346x_{26} = 81.6814089933346
x27=−6.28318530717959x_{27} = -6.28318530717959
x28=96.7610537305656x_{28} = 96.7610537305656
x29=−51.5221195188726x_{29} = -51.5221195188726
x30=23.8761041672824x_{30} = 23.8761041672824
x31=31.4159265358979x_{31} = 31.4159265358979
x32=94.2477796076938x_{32} = 94.2477796076938
x33=13.8230076757951x_{33} = 13.8230076757951
x34=−54.0353936417444x_{34} = -54.0353936417444
x35=−46.4955712731289x_{35} = -46.4955712731289
x36=−11.3097335529233x_{36} = -11.3097335529233
x37=74.1415866247191x_{37} = 74.1415866247191
x38=21.3628300444106x_{38} = 21.3628300444106
x39=54.0353936417444x_{39} = 54.0353936417444
x40=18.8495559215388x_{40} = 18.8495559215388
x41=−16.3362817986669x_{41} = -16.3362817986669
x42=84.1946831162065x_{42} = 84.1946831162065
x43=−26.3893782901543x_{43} = -26.3893782901543
x44=−23.8761041672824x_{44} = -23.8761041672824
x45=−59.0619418874881x_{45} = -59.0619418874881
x46=−79.1681348704628x_{46} = -79.1681348704628
x47=89.2212313619501x_{47} = 89.2212313619501
x48=8.79645943005142x_{48} = 8.79645943005142
x49=11.3097335529233x_{49} = 11.3097335529233
x50=99.2743278534375x_{50} = 99.2743278534375
x51=−8.79645943005142x_{51} = -8.79645943005142
x52=46.4955712731289x_{52} = 46.4955712731289
x53=16.3362817986669x_{53} = 16.3362817986669
x54=−56.5486677646163x_{54} = -56.5486677646163
x55=−43.9822971502571x_{55} = -43.9822971502571
x56=−86.7079572390783x_{56} = -86.7079572390783
x57=−13.8230076757951x_{57} = -13.8230076757951
x58=−18.8495559215388x_{58} = -18.8495559215388
x59=−3.76991118430775x_{59} = -3.76991118430775
x60=33.9292006587698x_{60} = 33.9292006587698
x61=91.734505484822x_{61} = 91.734505484822
x62=64.0884901332318x_{62} = 64.0884901332318
x63=61.5752160103599x_{63} = 61.5752160103599
x64=−31.4159265358979x_{64} = -31.4159265358979
x65=−28.9026524130261x_{65} = -28.9026524130261
x66=79.1681348704628x_{66} = 79.1681348704628
x67=−84.1946831162065x_{67} = -84.1946831162065
x68=59.0619418874881x_{68} = 59.0619418874881
x69=66.6017642561036x_{69} = 66.6017642561036
x70=26.3893782901543x_{70} = 26.3893782901543
x71=56.5486677646163x_{71} = 56.5486677646163
x72=−49.0088453960008x_{72} = -49.0088453960008
x73=38.9557489045134x_{73} = 38.9557489045134
x74=49.0088453960008x_{74} = 49.0088453960008
Extremos de la función
Para hallar los extremos hay que resolver la ecuación
ddxf(x)=0\frac{d}{d x} f{\left(x \right)} = 0
(la derivada es igual a cero),
y las raíces de esta ecuación serán los extremos de esta función:
ddxf(x)=\frac{d}{d x} f{\left(x \right)} =
primera derivada
tan⁡2(x)+cot⁡2(x4)4+54=0\tan^{2}{\left(x \right)} + \frac{\cot^{2}{\left(\frac{x}{4} \right)}}{4} + \frac{5}{4} = 0
Resolvermos esta ecuación
Soluciones no halladas,
tal vez la función no tenga extremos
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→−∞(tan⁡(x)−cot⁡(x4))y = \lim_{x \to -\infty}\left(\tan{\left(x \right)} - \cot{\left(\frac{x}{4} \right)}\right)
True

Tomamos como el límite
es decir,
ecuación de la asíntota horizontal a la derecha:
y=lim⁡x→∞(tan⁡(x)−cot⁡(x4))y = \lim_{x \to \infty}\left(\tan{\left(x \right)} - \cot{\left(\frac{x}{4} \right)}\right)
Asíntotas inclinadas
Se puede hallar la asíntota inclinada calculando el límite de la función tan(x) - cot(x/4), 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→−∞(tan⁡(x)−cot⁡(x4)x)y = x \lim_{x \to -\infty}\left(\frac{\tan{\left(x \right)} - \cot{\left(\frac{x}{4} \right)}}{x}\right)
True

Tomamos como el límite
es decir,
ecuación de la asíntota inclinada a la derecha:
y=xlim⁡x→∞(tan⁡(x)−cot⁡(x4)x)y = x \lim_{x \to \infty}\left(\frac{\tan{\left(x \right)} - \cot{\left(\frac{x}{4} \right)}}{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:
tan⁡(x)−cot⁡(x4)=−tan⁡(x)+cot⁡(x4)\tan{\left(x \right)} - \cot{\left(\frac{x}{4} \right)} = - \tan{\left(x \right)} + \cot{\left(\frac{x}{4} \right)}
- No
tan⁡(x)−cot⁡(x4)=tan⁡(x)−cot⁡(x4)\tan{\left(x \right)} - \cot{\left(\frac{x}{4} \right)} = \tan{\left(x \right)} - \cot{\left(\frac{x}{4} \right)}
- No
es decir, función
no es
par ni impar
Gráfico
Gráfico de la función y = y=tg(x)-ctg(x/4)
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