Mister Exam

Graphing y = exp(arctgx)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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        acot(x)
f(x) = e       
$$f{\left(x \right)} = e^{\operatorname{acot}{\left(x \right)}}$$
f = exp(acot(x))
The graph of the function
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:
$$e^{\operatorname{acot}{\left(x \right)}} = 0$$
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 exp(acot(x)).
$$e^{\operatorname{acot}{\left(0 \right)}}$$
The result:
$$f{\left(0 \right)} = e^{\frac{\pi}{2}}$$
The point:
(0, exp(pi/2))
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
$$- \frac{e^{\operatorname{acot}{\left(x \right)}}}{x^{2} + 1} = 0$$
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
$$\frac{\left(2 x + 1\right) e^{\operatorname{acot}{\left(x \right)}}}{\left(x^{2} + 1\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{1}{2}$$

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
$$\left[- \frac{1}{2}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, - \frac{1}{2}\right]$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} e^{\operatorname{acot}{\left(x \right)}} = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty} e^{\operatorname{acot}{\left(x \right)}} = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of exp(acot(x)), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{e^{\operatorname{acot}{\left(x \right)}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{e^{\operatorname{acot}{\left(x \right)}}}{x}\right) = 0$$
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:
$$e^{\operatorname{acot}{\left(x \right)}} = e^{- \operatorname{acot}{\left(x \right)}}$$
- No
$$e^{\operatorname{acot}{\left(x \right)}} = - e^{- \operatorname{acot}{\left(x \right)}}$$
- No
so, the function
not is
neither even, nor odd
The graph
Graphing y = exp(arctgx)