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Trazado el gráfico de la función/ (2*cos(x)+1)/sqrt(-x^2+6*x+27)

Gráfico de la función y = (2*cos(x)+1)/sqrt(-x^2+6*x+27)

v

Gráfico:

interior superior

Puntos de intersección:

mostrar?

Definida a trozos:

Solución

Ha introducido [src] 
           2*cos(x) + 1    
f(x) = --------------------
          _________________
         /    2            
       \/  - x  + 6*x + 27
f(x)=2cos⁡(x)+1(−x2+6x)+27f{\left(x \right)} = \frac{2 \cos{\left(x \right)} + 1}{\sqrt{\left(- x^{2} + 6 x\right) + 27}} 
f = (2*cos(x) + 1)/sqrt(-x^2 + 6*x + 27)
Gráfico de la función
02468-8-6-4-2-10105-5
Dominio de definición de la función
Puntos en los que la función no está definida exactamente:
x1=−3x_{1} = -3
x2=9x_{2} = 9
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:
2cos⁡(x)+1(−x2+6x)+27=0\frac{2 \cos{\left(x \right)} + 1}{\sqrt{\left(- x^{2} + 6 x\right) + 27}} = 0
Resolvermos esta ecuación
Puntos de cruce con el eje X:

Solución analítica
x1=2π3x_{1} = \frac{2 \pi}{3}
x2=4π3x_{2} = \frac{4 \pi}{3}
Solución numérica
x1=16.7551608191456x_{1} = 16.7551608191456
x2=−77.4926187885482x_{2} = -77.4926187885482
x3=−14.6607657167524x_{3} = -14.6607657167524
x4=−102.625360017267x_{4} = -102.625360017267
x5=46.0766922526503x_{5} = 46.0766922526503
x6=−85.870199198121x_{6} = -85.870199198121
x7=−4.18879020478639x_{7} = -4.18879020478639
x8=96.342174710087x_{8} = 96.342174710087
x9=29.3215314335047x_{9} = 29.3215314335047
x10=4.18879020478639x_{10} = 4.18879020478639
x11=−73.3038285837618x_{11} = -73.3038285837618
x12=−71.2094334813686x_{12} = -71.2094334813686
x13=−2.0943951023932x_{13} = -2.0943951023932
x14=284.837733925475x_{14} = 284.837733925475
x15=73.3038285837618x_{15} = 73.3038285837618
x16=8.37758040957278x_{16} = 8.37758040957278
x17=64.9262481741891x_{17} = 64.9262481741891
x18=−96.342174710087x_{18} = -96.342174710087
x19=41.8879020478639x_{19} = 41.8879020478639
x20=39.7935069454707x_{20} = 39.7935069454707
x21=−92.1533845053006x_{21} = -92.1533845053006
x22=27.2271363311115x_{22} = 27.2271363311115
x23=−29.3215314335047x_{23} = -29.3215314335047
x24=23.0383461263252x_{24} = 23.0383461263252
x25=20.943951023932x_{25} = 20.943951023932
x26=54.4542726622231x_{26} = 54.4542726622231
x27=2.0943951023932x_{27} = 2.0943951023932
x28=10.471975511966x_{28} = 10.471975511966
x29=60.7374579694027x_{29} = 60.7374579694027
x30=−39.7935069454707x_{30} = -39.7935069454707
x31=−46.0766922526503x_{31} = -46.0766922526503
x32=−64.9262481741891x_{32} = -64.9262481741891
x33=−27.2271363311115x_{33} = -27.2271363311115
x34=−90.0589894029074x_{34} = -90.0589894029074
x35=−52.3598775598299x_{35} = -52.3598775598299
x36=48.1710873550435x_{36} = 48.1710873550435
x37=−41.8879020478639x_{37} = -41.8879020478639
x38=67.0206432765823x_{38} = 67.0206432765823
x39=71.2094334813686x_{39} = 71.2094334813686
x40=−79.5870138909414x_{40} = -79.5870138909414
x41=−58.6430628670095x_{41} = -58.6430628670095
x42=58.6430628670095x_{42} = 58.6430628670095
x43=90.0589894029074x_{43} = 90.0589894029074
x44=77.4926187885482x_{44} = 77.4926187885482
x45=−7786.96099070271x_{45} = -7786.96099070271
x46=−33.5103216382911x_{46} = -33.5103216382911
x47=85.870199198121x_{47} = 85.870199198121
x48=35.6047167406843x_{48} = 35.6047167406843
x49=−98.4365698124802x_{49} = -98.4365698124802
x50=−23.0383461263252x_{50} = -23.0383461263252
x51=−83.7758040957278x_{51} = -83.7758040957278
x52=−35.6047167406843x_{52} = -35.6047167406843
x53=−16.7551608191456x_{53} = -16.7551608191456
x54=92.1533845053006x_{54} = 92.1533845053006
x55=83.7758040957278x_{55} = 83.7758040957278
x56=14.6607657167524x_{56} = 14.6607657167524
x57=−67.0206432765823x_{57} = -67.0206432765823
x58=−54.4542726622231x_{58} = -54.4542726622231
x59=33.5103216382911x_{59} = 33.5103216382911
x60=−8.37758040957278x_{60} = -8.37758040957278
x61=−48.1710873550435x_{61} = -48.1710873550435
x62=52.3598775598299x_{62} = 52.3598775598299
x63=79.5870138909414x_{63} = 79.5870138909414
x64=−10.471975511966x_{64} = -10.471975511966
x65=−20.943951023932x_{65} = -20.943951023932
x66=98.4365698124802x_{66} = 98.4365698124802
x67=−60.7374579694027x_{67} = -60.7374579694027
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 (2*cos(x) + 1)/sqrt(-x^2 + 6*x + 27).
1+2cos⁡(0)(−02+0⋅6)+27\frac{1 + 2 \cos{\left(0 \right)}}{\sqrt{\left(- 0^{2} + 0 \cdot 6\right) + 27}}
Resultado:
f(0)=33f{\left(0 \right)} = \frac{\sqrt{3}}{3}
Punto:
(0, sqrt(3)/3)
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
−(3−x)(2cos⁡(x)+1)((−x2+6x)+27)32−2sin⁡(x)(−x2+6x)+27=0- \frac{\left(3 - x\right) \left(2 \cos{\left(x \right)} + 1\right)}{\left(\left(- x^{2} + 6 x\right) + 27\right)^{\frac{3}{2}}} - \frac{2 \sin{\left(x \right)}}{\sqrt{\left(- x^{2} + 6 x\right) + 27}} = 0
Resolvermos esta ecuación
Soluciones no halladas,
tal vez la función no tenga extremos
Puntos de flexiones
Hallemos los puntos de flexiones, para eso hay que resolver la ecuación
d2dx2f(x)=0\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0
(la segunda derivada es igual a cero),
las raíces de la ecuación obtenida serán los puntos de flexión para el gráfico de la función indicado:
d2dx2f(x)=\frac{d^{2}}{d x^{2}} f{\left(x \right)} =
segunda derivada
−4(x−3)sin⁡(x)−x2+6x+27+(3(x−3)2−x2+6x+27+1)(2cos⁡(x)+1)−x2+6x+27−2cos⁡(x)−x2+6x+27=0\frac{- \frac{4 \left(x - 3\right) \sin{\left(x \right)}}{- x^{2} + 6 x + 27} + \frac{\left(\frac{3 \left(x - 3\right)^{2}}{- x^{2} + 6 x + 27} + 1\right) \left(2 \cos{\left(x \right)} + 1\right)}{- x^{2} + 6 x + 27} - 2 \cos{\left(x \right)}}{\sqrt{- x^{2} + 6 x + 27}} = 0
Resolvermos esta ecuación
Raíces de esta ecuación
x1=13.8465323761669x_{1} = 13.8465323761669
x2=73.7991733347854x_{2} = 73.7991733347854
x3=58.0823708717371x_{3} = 58.0823708717371
x4=−20.326218079642x_{4} = -20.326218079642
x5=80.0846768988467x_{5} = 80.0846768988467
x6=−10.8248348359704x_{6} = -10.8248348359704
x7=−32.9285961487609x_{7} = -32.9285961487609
x8=39.2122701598178x_{8} = 39.2122701598178
x9=−36.0765683290123x_{9} = -36.0765683290123
x10=−7.54517548183453x_{10} = -7.54517548183453
x11=98.9393484060984x_{11} = 98.9393484060984
x12=−67.5158736689866x_{12} = -67.5158736689866
x13=−61.2298870085564x_{13} = -61.2298870085564
x14=−95.7981529014901x_{14} = -95.7981529014901
x15=42.3601393138239x_{15} = 42.3601393138239
x16=−17.1728414519822x_{16} = -17.1728414519822
x17=70.6558116942595x_{17} = 70.6558116942595
x18=−76.9436993160132x_{18} = -76.9436993160132
x19=−70.6583100968917x_{19} = -70.6583100968917
x20=92.654699035415x_{20} = 92.654699035415
x21=83.2269756234019x_{21} = 83.2269756234019
x22=−13.9974751167387x_{22} = -13.9974751167387
x23=67.5132105338142x_{23} = 67.5132105338142
x24=−89.5135610668403x_{24} = -89.5135610668403
x25=86.3698269774228x_{25} = 86.3698269774228
x26=−45.5107709254369x_{26} = -45.5107709254369
x27=61.2266354014082x_{27} = 61.2266354014082
x28=36.0667238438788x_{28} = 36.0667238438788
x29=76.9416030161119x_{29} = 76.9416030161119
x30=−64.3724971177503x_{30} = -64.3724971177503
x31=64.3694674005225x_{31} = 64.3694674005225
x32=−48.6558162249209x_{32} = -48.6558162249209
x33=32.915769423701x_{33} = 32.915769423701
x34=−26.6318315604662x_{34} = -26.6318315604662
x35=29.7679605676882x_{35} = 29.7679605676882
x36=1.4572511255652x_{36} = 1.4572511255652
x37=−39.2209523298552x_{37} = -39.2209523298552
x38=−54.9432795904147x_{38} = -54.9432795904147
x39=17.1128566973379x_{39} = 17.1128566973379
x40=−98.9405769017429x_{40} = -98.9405769017429
x41=−92.6561012824251x_{41} = -92.6561012824251
x42=−29.7829773808165x_{42} = -29.7829773808165
x43=−83.2287600995735x_{43} = -83.2287600995735
x44=23.4577360550277x_{44} = 23.4577360550277
x45=−124.07719780293x_{45} = -124.07719780293
x46=−42.3671234966496x_{46} = -42.3671234966496
x47=−51.7989871798489x_{47} = -51.7989871798489
x48=−80.0865595804462x_{48} = -80.0865595804462
x49=−58.0861238269148x_{49} = -58.0861238269148
x50=4.85118758551886x_{50} = 4.85118758551886
x51=26.6107304927093x_{51} = 26.6107304927093
x52=48.6505909875265x_{52} = 48.6505909875265
x53=45.5044819454961x_{53} = 45.5044819454961
x54=95.7968139412703x_{54} = 95.7968139412703
x55=−86.3714428546585x_{55} = -86.3714428546585
x56=−23.4838760836324x_{56} = -23.4838760836324
x57=51.7942124700461x_{57} = 51.7942124700461
x58=54.939217302445x_{58} = 54.939217302445
x59=20.2837437320285x_{59} = 20.2837437320285
x60=89.5120233806755x_{60} = 89.5120233806755
x61=−359.706835597069x_{61} = -359.706835597069
x62=−73.8013953765254x_{62} = -73.8013953765254
Además hay que calcular los límites de y'' para los argumentos tendientes a los puntos de indeterminación de la función:
Puntos donde hay indeterminación:
x1=−3x_{1} = -3
x2=9x_{2} = 9

lim⁡x→−3−(−4(x−3)sin⁡(x)−x2+6x+27+(3(x−3)2−x2+6x+27+1)(2cos⁡(x)+1)−x2+6x+27−2cos⁡(x)−x2+6x+27)=∞i\lim_{x \to -3^-}\left(\frac{- \frac{4 \left(x - 3\right) \sin{\left(x \right)}}{- x^{2} + 6 x + 27} + \frac{\left(\frac{3 \left(x - 3\right)^{2}}{- x^{2} + 6 x + 27} + 1\right) \left(2 \cos{\left(x \right)} + 1\right)}{- x^{2} + 6 x + 27} - 2 \cos{\left(x \right)}}{\sqrt{- x^{2} + 6 x + 27}}\right) = \infty i
lim⁡x→−3+(−4(x−3)sin⁡(x)−x2+6x+27+(3(x−3)2−x2+6x+27+1)(2cos⁡(x)+1)−x2+6x+27−2cos⁡(x)−x2+6x+27)=−∞\lim_{x \to -3^+}\left(\frac{- \frac{4 \left(x - 3\right) \sin{\left(x \right)}}{- x^{2} + 6 x + 27} + \frac{\left(\frac{3 \left(x - 3\right)^{2}}{- x^{2} + 6 x + 27} + 1\right) \left(2 \cos{\left(x \right)} + 1\right)}{- x^{2} + 6 x + 27} - 2 \cos{\left(x \right)}}{\sqrt{- x^{2} + 6 x + 27}}\right) = -\infty
- los límites no son iguales, signo
x1=−3x_{1} = -3
- es el punto de flexión
lim⁡x→9−(−4(x−3)sin⁡(x)−x2+6x+27+(3(x−3)2−x2+6x+27+1)(2cos⁡(x)+1)−x2+6x+27−2cos⁡(x)−x2+6x+27)=−∞\lim_{x \to 9^-}\left(\frac{- \frac{4 \left(x - 3\right) \sin{\left(x \right)}}{- x^{2} + 6 x + 27} + \frac{\left(\frac{3 \left(x - 3\right)^{2}}{- x^{2} + 6 x + 27} + 1\right) \left(2 \cos{\left(x \right)} + 1\right)}{- x^{2} + 6 x + 27} - 2 \cos{\left(x \right)}}{\sqrt{- x^{2} + 6 x + 27}}\right) = -\infty
lim⁡x→9+(−4(x−3)sin⁡(x)−x2+6x+27+(3(x−3)2−x2+6x+27+1)(2cos⁡(x)+1)−x2+6x+27−2cos⁡(x)−x2+6x+27)=∞i\lim_{x \to 9^+}\left(\frac{- \frac{4 \left(x - 3\right) \sin{\left(x \right)}}{- x^{2} + 6 x + 27} + \frac{\left(\frac{3 \left(x - 3\right)^{2}}{- x^{2} + 6 x + 27} + 1\right) \left(2 \cos{\left(x \right)} + 1\right)}{- x^{2} + 6 x + 27} - 2 \cos{\left(x \right)}}{\sqrt{- x^{2} + 6 x + 27}}\right) = \infty i
- los límites no son iguales, signo
x2=9x_{2} = 9
- es el punto de flexión

Intervalos de convexidad y concavidad:
Hallemos los intervales donde la función es convexa o cóncava, para eso veamos cómo se comporta la función en los puntos de flexiones:
Cóncava en los intervalos
[1.4572511255652,4.85118758551886]\left[1.4572511255652, 4.85118758551886\right]
Convexa en los intervalos
(−∞,1.4572511255652]∪[4.85118758551886,∞)\left(-\infty, 1.4572511255652\right] \cup \left[4.85118758551886, \infty\right)
Asíntotas verticales
Hay:
x1=−3x_{1} = -3
x2=9x_{2} = 9
Asíntotas horizontales
Hallemos las asíntotas horizontales mediante los límites de esta función con x->+oo y x->-oo
lim⁡x→−∞(2cos⁡(x)+1(−x2+6x)+27)=0\lim_{x \to -\infty}\left(\frac{2 \cos{\left(x \right)} + 1}{\sqrt{\left(- x^{2} + 6 x\right) + 27}}\right) = 0
Tomamos como el límite
es decir,
ecuación de la asíntota horizontal a la izquierda:
y=0y = 0
lim⁡x→∞(2cos⁡(x)+1(−x2+6x)+27)=0\lim_{x \to \infty}\left(\frac{2 \cos{\left(x \right)} + 1}{\sqrt{\left(- x^{2} + 6 x\right) + 27}}\right) = 0
Tomamos como el límite
es decir,
ecuación de la asíntota horizontal a la derecha:
y=0y = 0
Asíntotas inclinadas
Se puede hallar la asíntota inclinada calculando el límite de la función (2*cos(x) + 1)/sqrt(-x^2 + 6*x + 27), dividida por x con x->+oo y x ->-oo
lim⁡x→−∞(2cos⁡(x)+1x(−x2+6x)+27)=0\lim_{x \to -\infty}\left(\frac{2 \cos{\left(x \right)} + 1}{x \sqrt{\left(- x^{2} + 6 x\right) + 27}}\right) = 0
Tomamos como el límite
es decir,
la inclinada coincide con la asíntota horizontal a la derecha
lim⁡x→∞(2cos⁡(x)+1x(−x2+6x)+27)=0\lim_{x \to \infty}\left(\frac{2 \cos{\left(x \right)} + 1}{x \sqrt{\left(- x^{2} + 6 x\right) + 27}}\right) = 0
Tomamos como el límite
es decir,
la inclinada coincide con la asíntota horizontal a la izquierda
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:
2cos⁡(x)+1(−x2+6x)+27=2cos⁡(x)+1−x2−6x+27\frac{2 \cos{\left(x \right)} + 1}{\sqrt{\left(- x^{2} + 6 x\right) + 27}} = \frac{2 \cos{\left(x \right)} + 1}{\sqrt{- x^{2} - 6 x + 27}}
- No
2cos⁡(x)+1(−x2+6x)+27=−2cos⁡(x)+1−x2−6x+27\frac{2 \cos{\left(x \right)} + 1}{\sqrt{\left(- x^{2} + 6 x\right) + 27}} = - \frac{2 \cos{\left(x \right)} + 1}{\sqrt{- x^{2} - 6 x + 27}}
- No
es decir, función
no es
par ni impar
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