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Plotting a function graph/ (1+x)^(1/x)

Graphing y = (1+x)^(1/x)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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       x _______
f(x) = \/ 1 + x
f(x)=(x+1)1xf{\left(x \right)} = \left(x + 1\right)^{\frac{1}{x}} 
f = (x + 1)^(1/x)
The graph of the function
-1.0-0.55.00.00.51.01.52.02.53.03.54.04.502000
The domain of the function
The points at which the function is not precisely defined:
x1=0x_{1} = 0
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis X at f = 0
so we need to solve the equation:
(x+1)1x=0\left(x + 1\right)^{\frac{1}{x}} = 0
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (1 + x)^(1/x).
1101^{\frac{1}{0}}
The result:
f(0)=NaNf{\left(0 \right)} = \text{NaN}
- the solutions of the equation d'not exist
Extrema of the function
In order to find the extrema, we need to solve the equation
ddxf(x)=0\frac{d}{d x} f{\left(x \right)} = 0
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
ddxf(x)=\frac{d}{d x} f{\left(x \right)} =
the first derivative
(x+1)1x(1x(x+1)−log⁡(x+1)x2)=0\left(x + 1\right)^{\frac{1}{x}} \left(\frac{1}{x \left(x + 1\right)} - \frac{\log{\left(x + 1 \right)}}{x^{2}}\right) = 0
Solve this equation
Solutions are not found,
function may have no extrema
Inflection points
Let's find the inflection points, we'll need to solve the equation for this
d2dx2f(x)=0\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
d2dx2f(x)=\frac{d^{2}}{d x^{2}} f{\left(x \right)} =
the second derivative
(x+1)1x(−1(x+1)2+(1x+1−log⁡(x+1)x)2x−2x(x+1)+2log⁡(x+1)x2)x=0\frac{\left(x + 1\right)^{\frac{1}{x}} \left(- \frac{1}{\left(x + 1\right)^{2}} + \frac{\left(\frac{1}{x + 1} - \frac{\log{\left(x + 1 \right)}}{x}\right)^{2}}{x} - \frac{2}{x \left(x + 1\right)} + \frac{2 \log{\left(x + 1 \right)}}{x^{2}}\right)}{x} = 0
Solve this equation
The roots of this equation
x1=29980.4129184309x_{1} = 29980.4129184309
x2=46714.933356596x_{2} = 46714.933356596
x3=25461.2257552751x_{3} = 25461.2257552751
x4=47821.3520867522x_{4} = 47821.3520867522
x5=24326.4600582814x_{5} = 24326.4600582814
x6=54440.8206019301x_{6} = 54440.8206019301
x7=35591.3063506813x_{7} = 35591.3063506813
x8=34472.1244637095x_{8} = 34472.1244637095
x9=28853.3975120921x_{9} = 28853.3975120921
x10=48926.8209166782x_{10} = 48926.8209166782
x11=37825.5858265746x_{11} = 37825.5858265746
x12=43389.6836739697x_{12} = 43389.6836739697
x13=27724.5899421305x_{13} = 27724.5899421305
x14=32229.39409249x_{14} = 32229.39409249
x15=50031.3669817923x_{15} = 50031.3669817923
x16=44499.1301965023x_{16} = 44499.1301965023
x17=38940.7849760761x_{17} = 38940.7849760761
x18=31105.7195519117x_{18} = 31105.7195519117
x19=42279.1627921657x_{19} = 42279.1627921657
x20=45607.5361227612x_{20} = 45607.5361227612
x21=53339.7200843891x_{21} = 53339.7200843891
x22=33351.507260858x_{22} = 33351.507260858
x23=52237.7926464356x_{23} = 52237.7926464356
x24=40054.7526229321x_{24} = 40054.7526229321
x25=41167.5317686861x_{25} = 41167.5317686861
x26=51135.0160556731x_{26} = 51135.0160556731
x27=36709.1092974302x_{27} = 36709.1092974302
x28=26593.8992266672x_{28} = 26593.8992266672
x29=55541.1154065021x_{29} = 55541.1154065021
x30=56640.6247473501x_{30} = 56640.6247473501
x31=57739.3679767074x_{31} = 57739.3679767074
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
x1=0x_{1} = 0

lim⁡x→0−((x+1)1x(−1(x+1)2+(1x+1−log⁡(x+1)x)2x−2x(x+1)+2log⁡(x+1)x2)x)=11e12\lim_{x \to 0^-}\left(\frac{\left(x + 1\right)^{\frac{1}{x}} \left(- \frac{1}{\left(x + 1\right)^{2}} + \frac{\left(\frac{1}{x + 1} - \frac{\log{\left(x + 1 \right)}}{x}\right)^{2}}{x} - \frac{2}{x \left(x + 1\right)} + \frac{2 \log{\left(x + 1 \right)}}{x^{2}}\right)}{x}\right) = \frac{11 e}{12}
lim⁡x→0+((x+1)1x(−1(x+1)2+(1x+1−log⁡(x+1)x)2x−2x(x+1)+2log⁡(x+1)x2)x)=11e12\lim_{x \to 0^+}\left(\frac{\left(x + 1\right)^{\frac{1}{x}} \left(- \frac{1}{\left(x + 1\right)^{2}} + \frac{\left(\frac{1}{x + 1} - \frac{\log{\left(x + 1 \right)}}{x}\right)^{2}}{x} - \frac{2}{x \left(x + 1\right)} + \frac{2 \log{\left(x + 1 \right)}}{x^{2}}\right)}{x}\right) = \frac{11 e}{12}
- limits are equal, then skip the corresponding point

Сonvexity and concavity intervals:
Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Have no bends at the whole real axis
Vertical asymptotes
Have:
x1=0x_{1} = 0
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(x+1)1x=1\lim_{x \to -\infty} \left(x + 1\right)^{\frac{1}{x}} = 1
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=1y = 1
lim⁡x→∞(x+1)1x=1\lim_{x \to \infty} \left(x + 1\right)^{\frac{1}{x}} = 1
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=1y = 1
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (1 + x)^(1/x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞((x+1)1xx)=0\lim_{x \to -\infty}\left(\frac{\left(x + 1\right)^{\frac{1}{x}}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞((x+1)1xx)=0\lim_{x \to \infty}\left(\frac{\left(x + 1\right)^{\frac{1}{x}}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
Even and odd functions
Let's check, whether the function even or odd by using relations f = f(-x) и f = -f(-x).
So, check:
(x+1)1x=(1−x)−1x\left(x + 1\right)^{\frac{1}{x}} = \left(1 - x\right)^{- \frac{1}{x}}
- No
(x+1)1x=−(1−x)−1x\left(x + 1\right)^{\frac{1}{x}} = - \left(1 - x\right)^{- \frac{1}{x}}
- No
so, the function
not is
neither even, nor odd
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