Mister Exam
Graphing y = e^(1/(2^x-1))
The solution
You have entered
[src]
1
------
x
2 - 1
f(x) = E
$$f{\left(x \right)} = e^{\frac{1}{2^{x} - 1}}$$
f = E^(1/(2^x - 1))
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 0$$
$$x_{1} = 0$$
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:
$$e^{\frac{1}{2^{x} - 1}} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
so we need to solve the equation:
$$e^{\frac{1}{2^{x} - 1}} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
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
$$- \frac{2^{x} e^{\frac{1}{2^{x} - 1}} \log{\left(2 \right)}}{\left(2^{x} - 1\right)^{2}} = 0$$
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
$$- \frac{2^{x} e^{\frac{1}{2^{x} - 1}} \log{\left(2 \right)}}{\left(2^{x} - 1\right)^{2}} = 0$$
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
$$\frac{2^{x} \left(\frac{2 \cdot 2^{x}}{2^{x} - 1} + \frac{2^{x}}{\left(2^{x} - 1\right)^{2}} - 1\right) e^{\frac{1}{2^{x} - 1}} \log{\left(2 \right)}^{2}}{\left(2^{x} - 1\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = -1 + \frac{\log{\left(-1 + \sqrt{5} \right)}}{\log{\left(2 \right)}}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(\frac{2^{x} \left(\frac{2 \cdot 2^{x}}{2^{x} - 1} + \frac{2^{x}}{\left(2^{x} - 1\right)^{2}} - 1\right) e^{\frac{1}{2^{x} - 1}} \log{\left(2 \right)}^{2}}{\left(2^{x} - 1\right)^{2}}\right) = 0$$
$$\lim_{x \to 0^+}\left(\frac{2^{x} \left(\frac{2 \cdot 2^{x}}{2^{x} - 1} + \frac{2^{x}}{\left(2^{x} - 1\right)^{2}} - 1\right) e^{\frac{1}{2^{x} - 1}} \log{\left(2 \right)}^{2}}{\left(2^{x} - 1\right)^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = 0$$
- is an inflection 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:
Concave at the intervals
$$\left[-1 + \frac{\log{\left(-1 + \sqrt{5} \right)}}{\log{\left(2 \right)}}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, -1 + \frac{\log{\left(-1 + \sqrt{5} \right)}}{\log{\left(2 \right)}}\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{2^{x} \left(\frac{2 \cdot 2^{x}}{2^{x} - 1} + \frac{2^{x}}{\left(2^{x} - 1\right)^{2}} - 1\right) e^{\frac{1}{2^{x} - 1}} \log{\left(2 \right)}^{2}}{\left(2^{x} - 1\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = -1 + \frac{\log{\left(-1 + \sqrt{5} \right)}}{\log{\left(2 \right)}}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(\frac{2^{x} \left(\frac{2 \cdot 2^{x}}{2^{x} - 1} + \frac{2^{x}}{\left(2^{x} - 1\right)^{2}} - 1\right) e^{\frac{1}{2^{x} - 1}} \log{\left(2 \right)}^{2}}{\left(2^{x} - 1\right)^{2}}\right) = 0$$
$$\lim_{x \to 0^+}\left(\frac{2^{x} \left(\frac{2 \cdot 2^{x}}{2^{x} - 1} + \frac{2^{x}}{\left(2^{x} - 1\right)^{2}} - 1\right) e^{\frac{1}{2^{x} - 1}} \log{\left(2 \right)}^{2}}{\left(2^{x} - 1\right)^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = 0$$
- is an inflection 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:
Concave at the intervals
$$\left[-1 + \frac{\log{\left(-1 + \sqrt{5} \right)}}{\log{\left(2 \right)}}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, -1 + \frac{\log{\left(-1 + \sqrt{5} \right)}}{\log{\left(2 \right)}}\right]$$
Vertical asymptotes
Have:
$$x_{1} = 0$$
$$x_{1} = 0$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} e^{\frac{1}{2^{x} - 1}} = e^{-1}$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = e^{-1}$$
$$\lim_{x \to \infty} e^{\frac{1}{2^{x} - 1}} = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
$$\lim_{x \to -\infty} e^{\frac{1}{2^{x} - 1}} = e^{-1}$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = e^{-1}$$
$$\lim_{x \to \infty} e^{\frac{1}{2^{x} - 1}} = 1$$
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 E^(1/(2^x - 1)), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{e^{\frac{1}{2^{x} - 1}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{e^{\frac{1}{2^{x} - 1}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{e^{\frac{1}{2^{x} - 1}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{e^{\frac{1}{2^{x} - 1}}}{x}\right) = 0$$
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:
$$e^{\frac{1}{2^{x} - 1}} = e^{\frac{1}{-1 + 2^{- x}}}$$
- No
$$e^{\frac{1}{2^{x} - 1}} = - e^{\frac{1}{-1 + 2^{- x}}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$e^{\frac{1}{2^{x} - 1}} = e^{\frac{1}{-1 + 2^{- x}}}$$
- No
$$e^{\frac{1}{2^{x} - 1}} = - e^{\frac{1}{-1 + 2^{- x}}}$$
- No
so, the function
not is
neither even, nor odd