Mister Exam
Graphing y = x^e^x
The solution
The graph of the function
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} = $$
the first derivative
$$x^{e^{x}} \left(e^{x} \log{\left(x \right)} + \frac{e^{x}}{x}\right) = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} = $$
the first derivative
$$x^{e^{x}} \left(e^{x} \log{\left(x \right)} + \frac{e^{x}}{x}\right) = 0$$
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
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative
$$x^{e^{x}} \left(\left(\log{\left(x \right)} + \frac{1}{x}\right)^{2} e^{x} + \log{\left(x \right)} + \frac{2}{x} - \frac{1}{x^{2}}\right) e^{x} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 0.274972858182512$$
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
$$\left[0.274972858182512, \infty\right)$$
Convex at the intervals
$$\left(-\infty, 0.274972858182512\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative
$$x^{e^{x}} \left(\left(\log{\left(x \right)} + \frac{1}{x}\right)^{2} e^{x} + \log{\left(x \right)} + \frac{2}{x} - \frac{1}{x^{2}}\right) e^{x} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 0.274972858182512$$
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
$$\left[0.274972858182512, \infty\right)$$
Convex at the intervals
$$\left(-\infty, 0.274972858182512\right]$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} x^{e^{x}} = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty} x^{e^{x}} = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{x \to -\infty} x^{e^{x}} = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty} x^{e^{x}} = \infty$$
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
$$\lim_{x \to -\infty}\left(\frac{x^{e^{x}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{x^{e^{x}}}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\frac{x^{e^{x}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{x^{e^{x}}}{x}\right) = \infty$$
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:
$$x^{e^{x}} = \left(- x\right)^{e^{- x}}$$
- No
$$x^{e^{x}} = - \left(- x\right)^{e^{- x}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$x^{e^{x}} = \left(- x\right)^{e^{- x}}$$
- No
$$x^{e^{x}} = - \left(- x\right)^{e^{- x}}$$
- No
so, the function
not is
neither even, nor odd