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Plotting a function graph/ x^e^x

Graphing y = x^e^x

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
        / x\
        \E /
f(x) = x
f(x)=xexf{\left(x \right)} = x^{e^{x}} 
f = x^(E^x)
The graph of the function
02468-8-6-4-2-101002e307
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to x^(E^x).
0e00^{e^{0}}
The result:
f(0)=0f{\left(0 \right)} = 0
The point:
(0, 0)
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
xex(exlog⁡(x)+exx)=0x^{e^{x}} \left(e^{x} \log{\left(x \right)} + \frac{e^{x}}{x}\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
xex((log⁡(x)+1x)2ex+log⁡(x)+2x−1x2)ex=0x^{e^{x}} \left(\left(\log{\left(x \right)} + \frac{1}{x}\right)^{2} e^{x} + \log{\left(x \right)} + \frac{2}{x} - \frac{1}{x^{2}}\right) e^{x} = 0
Solve this equation
The roots of this equation
x1=0.274972858182512x_{1} = 0.274972858182512

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:
Concave at the intervals
[0.274972858182512,∞)\left[0.274972858182512, \infty\right)
Convex at the intervals
(−∞,0.274972858182512]\left(-\infty, 0.274972858182512\right]
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞xex=1\lim_{x \to -\infty} x^{e^{x}} = 1
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=1y = 1
lim⁡x→∞xex=∞\lim_{x \to \infty} x^{e^{x}} = \infty
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of x^(E^x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(xexx)=0\lim_{x \to -\infty}\left(\frac{x^{e^{x}}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(xexx)=∞\lim_{x \to \infty}\left(\frac{x^{e^{x}}}{x}\right) = \infty
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
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:
xex=(−x)e−xx^{e^{x}} = \left(- x\right)^{e^{- x}}
- No
xex=−(−x)e−xx^{e^{x}} = - \left(- x\right)^{e^{- x}}
- No
so, the function
not is
neither even, nor odd
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