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Trazado el gráfico de la función/ atan(5^x+1)^3

Gráfico de la función y = atan(5^x+1)^3

v

Gráfico:

interior superior

Puntos de intersección:

mostrar?

Definida a trozos:

Solución

Ha introducido [src] 
           3/ x    \
f(x) = atan \5  + 1/
f(x)=atan⁡3(5x+1)f{\left(x \right)} = \operatorname{atan}^{3}{\left(5^{x} + 1 \right)} 
f = atan(5^x + 1)^3
Gráfico de la función
02468-8-6-4-2-101005
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:
atan⁡3(5x+1)=0\operatorname{atan}^{3}{\left(5^{x} + 1 \right)} = 0
Resolvermos esta ecuación
Solución no hallada,
puede ser que el gráfico no cruce el eje X
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 atan(5^x + 1)^3.
atan⁡3(50+1)\operatorname{atan}^{3}{\left(5^{0} + 1 \right)}
Resultado:
f(0)=atan⁡3(2)f{\left(0 \right)} = \operatorname{atan}^{3}{\left(2 \right)}
Punto:
(0, atan(2)^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⋅5xlog⁡(5)atan⁡2(5x+1)(5x+1)2+1=0\frac{3 \cdot 5^{x} \log{\left(5 \right)} \operatorname{atan}^{2}{\left(5^{x} + 1 \right)}}{\left(5^{x} + 1\right)^{2} + 1} = 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
3⋅5x(−2⋅5x(5x+1)atan⁡(5x+1)(5x+1)2+1+2⋅5x(5x+1)2+1+atan⁡(5x+1))log⁡(5)2atan⁡(5x+1)(5x+1)2+1=0\frac{3 \cdot 5^{x} \left(- \frac{2 \cdot 5^{x} \left(5^{x} + 1\right) \operatorname{atan}{\left(5^{x} + 1 \right)}}{\left(5^{x} + 1\right)^{2} + 1} + \frac{2 \cdot 5^{x}}{\left(5^{x} + 1\right)^{2} + 1} + \operatorname{atan}{\left(5^{x} + 1 \right)}\right) \log{\left(5 \right)}^{2} \operatorname{atan}{\left(5^{x} + 1 \right)}}{\left(5^{x} + 1\right)^{2} + 1} = 0
Resolvermos esta ecuación
Raíces de esta ecuación
x1=−72.9148275920459x_{1} = -72.9148275920459
x2=91.2242768391714x_{2} = 91.2242768391714
x3=51.2242768391714x_{3} = 51.2242768391714
x4=−32.9148275920459x_{4} = -32.9148275920459
x5=−86.9148275920459x_{5} = -86.9148275920459
x6=−108.914827592046x_{6} = -108.914827592046
x7=103.224276839171x_{7} = 103.224276839171
x8=−30.9148275920457x_{8} = -30.9148275920457
x9=−58.9148275920459x_{9} = -58.9148275920459
x10=21.2242894893908x_{10} = 21.2242894893908
x11=−52.9148275920459x_{11} = -52.9148275920459
x12=63.2242768391714x_{12} = 63.2242768391714
x13=47.2242768391714x_{13} = 47.2242768391714
x14=−80.9148275920459x_{14} = -80.9148275920459
x15=41.2242768391714x_{15} = 41.2242768391714
x16=75.2242768391714x_{16} = 75.2242768391714
x17=−70.9148275920459x_{17} = -70.9148275920459
x18=−102.914827592046x_{18} = -102.914827592046
x19=−48.9148275920459x_{19} = -48.9148275920459
x20=65.2242768391714x_{20} = 65.2242768391714
x21=0.542176246771954x_{21} = 0.542176246771954
x22=39.2242768391714x_{22} = 39.2242768391714
x23=27.224276839981x_{23} = 27.224276839981
x24=−40.9148275920459x_{24} = -40.9148275920459
x25=83.2242768391714x_{25} = 83.2242768391714
x26=−34.9148275920459x_{26} = -34.9148275920459
x27=107.224276839171x_{27} = 107.224276839171
x28=−88.9148275920459x_{28} = -88.9148275920459
x29=−18.914784374042x_{29} = -18.914784374042
x30=25.2242768594116x_{30} = 25.2242768594116
x31=−74.9148275920459x_{31} = -74.9148275920459
x32=73.2242768391714x_{32} = 73.2242768391714
x33=−98.9148275920459x_{33} = -98.9148275920459
x34=35.2242768391714x_{34} = 35.2242768391714
x35=−90.9148275920459x_{35} = -90.9148275920459
x36=−36.9148275920459x_{36} = -36.9148275920459
x37=−54.9148275920459x_{37} = -54.9148275920459
x38=−42.9148275920459x_{38} = -42.9148275920459
x39=43.2242768391714x_{39} = 43.2242768391714
x40=93.2242768391714x_{40} = 93.2242768391714
x41=89.2242768391714x_{41} = 89.2242768391714
x42=−38.9148275920459x_{42} = -38.9148275920459
x43=57.2242768391714x_{43} = 57.2242768391714
x44=71.2242768391714x_{44} = 71.2242768391714
x45=−46.9148275920459x_{45} = -46.9148275920459
x46=95.2242768391714x_{46} = 95.2242768391714
x47=−26.9148275919352x_{47} = -26.9148275919352
x48=113.224276839171x_{48} = 113.224276839171
x49=55.2242768391714x_{49} = 55.2242768391714
x50=61.2242768391714x_{50} = 61.2242768391714
x51=33.2242768391715x_{51} = 33.2242768391715
x52=−44.9148275920459x_{52} = -44.9148275920459
x53=67.2242768391714x_{53} = 67.2242768391714
x54=87.2242768391714x_{54} = 87.2242768391714
x55=−62.9148275920459x_{55} = -62.9148275920459
x56=−106.914827592046x_{56} = -106.914827592046
x57=−100.914827592046x_{57} = -100.914827592046
x58=49.2242768391714x_{58} = 49.2242768391714
x59=99.2242768391714x_{59} = 99.2242768391714
x60=−68.9148275920459x_{60} = -68.9148275920459
x61=19.2245931342562x_{61} = 19.2245931342562
x62=59.2242768391714x_{62} = 59.2242768391714
x63=−82.9148275920459x_{63} = -82.9148275920459
x64=−20.9148258621738x_{64} = -20.9148258621738
x65=−94.9148275920459x_{65} = -94.9148275920459
x66=−92.9148275920459x_{66} = -92.9148275920459
x67=97.2242768391714x_{67} = 97.2242768391714
x68=−56.9148275920459x_{68} = -56.9148275920459
x69=−110.914827592046x_{69} = -110.914827592046
x70=31.2242768391727x_{70} = 31.2242768391727
x71=85.2242768391714x_{71} = 85.2242768391714
x72=45.2242768391714x_{72} = 45.2242768391714
x73=53.2242768391714x_{73} = 53.2242768391714
x74=−60.9148275920459x_{74} = -60.9148275920459
x75=−24.914827589278x_{75} = -24.914827589278
x76=77.2242768391714x_{76} = 77.2242768391714
x77=−22.9148275228492x_{77} = -22.9148275228492
x78=105.224276839171x_{78} = 105.224276839171
x79=69.2242768391714x_{79} = 69.2242768391714
x80=79.2242768391714x_{80} = 79.2242768391714
x81=−28.9148275920415x_{81} = -28.9148275920415
x82=−96.9148275920459x_{82} = -96.9148275920459
x83=23.2242773451776x_{83} = 23.2242773451776
x84=81.2242768391714x_{84} = 81.2242768391714
x85=−64.9148275920459x_{85} = -64.9148275920459
x86=−104.914827592046x_{86} = -104.914827592046
x87=−78.9148275920459x_{87} = -78.9148275920459
x88=101.224276839171x_{88} = 101.224276839171
x89=−66.9148275920459x_{89} = -66.9148275920459
x90=−50.9148275920459x_{90} = -50.9148275920459
x91=37.2242768391714x_{91} = 37.2242768391714
x92=109.224276839171x_{92} = 109.224276839171
x93=−84.9148275920459x_{93} = -84.9148275920459
x94=−112.914827592046x_{94} = -112.914827592046
x95=−76.9148275920459x_{95} = -76.9148275920459
x96=111.224276839171x_{96} = 111.224276839171
x97=29.2242768392038x_{97} = 29.2242768392038

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
(−∞,0.542176246771954]\left(-\infty, 0.542176246771954\right]
Convexa en los intervalos
[0.542176246771954,∞)\left[0.542176246771954, \infty\right)
Asíntotas horizontales
Hallemos las asíntotas horizontales mediante los límites de esta función con x->+oo y x->-oo
lim⁡x→−∞atan⁡3(5x+1)=π364\lim_{x \to -\infty} \operatorname{atan}^{3}{\left(5^{x} + 1 \right)} = \frac{\pi^{3}}{64}
Tomamos como el límite
es decir,
ecuación de la asíntota horizontal a la izquierda:
y=π364y = \frac{\pi^{3}}{64}
lim⁡x→∞atan⁡3(5x+1)=π38\lim_{x \to \infty} \operatorname{atan}^{3}{\left(5^{x} + 1 \right)} = \frac{\pi^{3}}{8}
Tomamos como el límite
es decir,
ecuación de la asíntota horizontal a la derecha:
y=π38y = \frac{\pi^{3}}{8}
Asíntotas inclinadas
Se puede hallar la asíntota inclinada calculando el límite de la función atan(5^x + 1)^3, dividida por x con x->+oo y x ->-oo
lim⁡x→−∞(atan⁡3(5x+1)x)=0\lim_{x \to -\infty}\left(\frac{\operatorname{atan}^{3}{\left(5^{x} + 1 \right)}}{x}\right) = 0
Tomamos como el límite
es decir,
la inclinada coincide con la asíntota horizontal a la derecha
lim⁡x→∞(atan⁡3(5x+1)x)=0\lim_{x \to \infty}\left(\frac{\operatorname{atan}^{3}{\left(5^{x} + 1 \right)}}{x}\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:
atan⁡3(5x+1)=atan⁡3(1+5−x)\operatorname{atan}^{3}{\left(5^{x} + 1 \right)} = \operatorname{atan}^{3}{\left(1 + 5^{- x} \right)}
- No
atan⁡3(5x+1)=−atan⁡3(1+5−x)\operatorname{atan}^{3}{\left(5^{x} + 1 \right)} = - \operatorname{atan}^{3}{\left(1 + 5^{- x} \right)}
- No
es decir, función
no es
par ni impar
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