Sr Examen

Otras calculadoras

Trazado el gráfico de la función/ (1-sqrt(1+abs(sin(x))))/(2+cos(x)*cos(x))

Gráfico de la función y = (1-sqrt(1+abs(sin(x))))/(2+cos(x)*cos(x))

v

Gráfico:

interior superior

Puntos de intersección:

mostrar?

Definida a trozos:

Solución

Ha introducido [src] 
             ______________
       1 - \/ 1 + |sin(x)| 
f(x) = --------------------
        2 + cos(x)*cos(x)
f(x)=1−∣sin⁡(x)∣+1cos⁡(x)cos⁡(x)+2f{\left(x \right)} = \frac{1 - \sqrt{\left|{\sin{\left(x \right)}}\right| + 1}}{\cos{\left(x \right)} \cos{\left(x \right)} + 2} 
f = (1 - sqrt(Abs(sin(x)) + 1))/(cos(x)*cos(x) + 2)
Gráfico de la función
02468-8-6-4-2-10100.25-0.25
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:
1−∣sin⁡(x)∣+1cos⁡(x)cos⁡(x)+2=0\frac{1 - \sqrt{\left|{\sin{\left(x \right)}}\right| + 1}}{\cos{\left(x \right)} \cos{\left(x \right)} + 2} = 0
Resolvermos esta ecuación
Puntos de cruce con el eje X:

Solución analítica
x1=0x_{1} = 0
x2=πx_{2} = \pi
Solución numérica
x1=69.1150383789755x_{1} = 69.1150383789755
x2=65.9734457253857x_{2} = 65.9734457253857
x3=−91.106186954104x_{3} = -91.106186954104
x4=−59.6902604182061x_{4} = -59.6902604182061
x5=−21.9911485751286x_{5} = -21.9911485751286
x6=12.5663706143592x_{6} = 12.5663706143592
x7=21.9911485751286x_{7} = 21.9911485751286
x8=−69.1150383789755x_{8} = -69.1150383789755
x9=−100.530964914873x_{9} = -100.530964914873
x10=3.14159265358979x_{10} = 3.14159265358979
x11=−3.14159265358979x_{11} = -3.14159265358979
x12=−25.1327412287183x_{12} = -25.1327412287183
x13=−15.707963267949x_{13} = -15.707963267949
x14=−53.4070751110265x_{14} = -53.4070751110265
x15=−72.2566310325652x_{15} = -72.2566310325652
x16=84.8230016469244x_{16} = 84.8230016469244
x17=−81.6814089933346x_{17} = -81.6814089933346
x18=−94.2477796076938x_{18} = -94.2477796076938
x19=18.8495559215388x_{19} = 18.8495559215388
x20=−65.9734457253857x_{20} = -65.9734457253857
x21=94.2477796076938x_{21} = 94.2477796076938
x22=9.42477796076938x_{22} = 9.42477796076938
x23=−40.8407044966673x_{23} = -40.8407044966673
x24=34.5575191894877x_{24} = 34.5575191894877
x25=0x_{25} = 0
x26=97.3893722612836x_{26} = 97.3893722612836
x27=53.4070751110265x_{27} = 53.4070751110265
x28=−62.8318530717959x_{28} = -62.8318530717959
x29=59.6902604182061x_{29} = 59.6902604182061
x30=−28.2743338823081x_{30} = -28.2743338823081
x31=−56.5486677646163x_{31} = -56.5486677646163
x32=91.106186954104x_{32} = 91.106186954104
x33=15.707963267949x_{33} = 15.707963267949
x34=−18.8495559215388x_{34} = -18.8495559215388
x35=6.28318530717959x_{35} = 6.28318530717959
x36=56.5486677646163x_{36} = 56.5486677646163
x37=9299.11425462579x_{37} = 9299.11425462579
x38=87.9645943005142x_{38} = 87.9645943005142
x39=31.4159265358979x_{39} = 31.4159265358979
x40=25.1327412287183x_{40} = 25.1327412287183
x41=43.9822971502571x_{41} = 43.9822971502571
x42=−47.1238898038469x_{42} = -47.1238898038469
x43=72.2566310325652x_{43} = 72.2566310325652
x44=−34.5575191894877x_{44} = -34.5575191894877
x45=−97.3893722612836x_{45} = -97.3893722612836
x46=−50.2654824574367x_{46} = -50.2654824574367
x47=100.530964914873x_{47} = 100.530964914873
x48=81.6814089933346x_{48} = 81.6814089933346
x49=−75.398223686155x_{49} = -75.398223686155
x50=40.8407044966673x_{50} = 40.8407044966673
x51=−9.42477796076938x_{51} = -9.42477796076938
x52=78.5398163397448x_{52} = 78.5398163397448
x53=−87.9645943005142x_{53} = -87.9645943005142
x54=37.6991118430775x_{54} = 37.6991118430775
x55=−78.5398163397448x_{55} = -78.5398163397448
x56=−6.28318530717959x_{56} = -6.28318530717959
x57=50.2654824574367x_{57} = 50.2654824574367
x58=−37.6991118430775x_{58} = -37.6991118430775
x59=−43.9822971502571x_{59} = -43.9822971502571
x60=47.1238898038469x_{60} = 47.1238898038469
x61=28.2743338823081x_{61} = 28.2743338823081
x62=62.8318530717959x_{62} = 62.8318530717959
x63=−31.4159265358979x_{63} = -31.4159265358979
x64=−12.5663706143592x_{64} = -12.5663706143592
x65=75.398223686155x_{65} = 75.398223686155
x66=−84.8230016469244x_{66} = -84.8230016469244
Puntos de cruce con el eje de coordenadas Y
El gráfico cruce el eje Y cuando x es igual a 0:
sustituimos x = 0 en (1 - sqrt(1 + Abs(sin(x))))/(2 + cos(x)*cos(x)).
1−∣sin⁡(0)∣+1cos⁡(0)cos⁡(0)+2\frac{1 - \sqrt{\left|{\sin{\left(0 \right)}}\right| + 1}}{\cos{\left(0 \right)} \cos{\left(0 \right)} + 2}
Resultado:
f(0)=0f{\left(0 \right)} = 0
Punto:
(0, 0)
Asíntotas horizontales
Hallemos las asíntotas horizontales mediante los límites de esta función con x->+oo y x->-oo
No se ha logrado calcular el límite a la izquierda
lim⁡x→−∞(1−∣sin⁡(x)∣+1cos⁡(x)cos⁡(x)+2)\lim_{x \to -\infty}\left(\frac{1 - \sqrt{\left|{\sin{\left(x \right)}}\right| + 1}}{\cos{\left(x \right)} \cos{\left(x \right)} + 2}\right)
No se ha logrado calcular el límite a la derecha
lim⁡x→∞(1−∣sin⁡(x)∣+1cos⁡(x)cos⁡(x)+2)\lim_{x \to \infty}\left(\frac{1 - \sqrt{\left|{\sin{\left(x \right)}}\right| + 1}}{\cos{\left(x \right)} \cos{\left(x \right)} + 2}\right)
Asíntotas inclinadas
Se puede hallar la asíntota inclinada calculando el límite de la función (1 - sqrt(1 + Abs(sin(x))))/(2 + cos(x)*cos(x)), dividida por x con x->+oo y x ->-oo
No se ha logrado calcular el límite a la izquierda
lim⁡x→−∞(1−∣sin⁡(x)∣+1x(cos⁡(x)cos⁡(x)+2))\lim_{x \to -\infty}\left(\frac{1 - \sqrt{\left|{\sin{\left(x \right)}}\right| + 1}}{x \left(\cos{\left(x \right)} \cos{\left(x \right)} + 2\right)}\right)
No se ha logrado calcular el límite a la derecha
lim⁡x→∞(1−∣sin⁡(x)∣+1x(cos⁡(x)cos⁡(x)+2))\lim_{x \to \infty}\left(\frac{1 - \sqrt{\left|{\sin{\left(x \right)}}\right| + 1}}{x \left(\cos{\left(x \right)} \cos{\left(x \right)} + 2\right)}\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:
1−∣sin⁡(x)∣+1cos⁡(x)cos⁡(x)+2=1−∣sin⁡(x)∣+1cos⁡(x)cos⁡(x)+2\frac{1 - \sqrt{\left|{\sin{\left(x \right)}}\right| + 1}}{\cos{\left(x \right)} \cos{\left(x \right)} + 2} = \frac{1 - \sqrt{\left|{\sin{\left(x \right)}}\right| + 1}}{\cos{\left(x \right)} \cos{\left(x \right)} + 2}
- Sí
1−∣sin⁡(x)∣+1cos⁡(x)cos⁡(x)+2=−1−∣sin⁡(x)∣+1cos⁡(x)cos⁡(x)+2\frac{1 - \sqrt{\left|{\sin{\left(x \right)}}\right| + 1}}{\cos{\left(x \right)} \cos{\left(x \right)} + 2} = - \frac{1 - \sqrt{\left|{\sin{\left(x \right)}}\right| + 1}}{\cos{\left(x \right)} \cos{\left(x \right)} + 2}
- No
es decir, función
es
par
    © Sr Examen – Калькулятор