Mister Exam
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-
How to use it?
- Graphing y =:
- x^3-5x
- x^3-4x^2+3x
- (x^3)/(4-x^2)
- x^3+3x-5
- Integral of d{x}:
-
e^x/(1+e^x)
- Derivative of:
-
e^x/(1+e^x)
- Limit of the function:
-
e^x/(1+e^x)
-
Identical expressions
- e^x/(one +e^x)
- e to the power of x divide by (1 plus e to the power of x)
- e to the power of x divide by (one plus e to the power of x)
- ex/(1+ex)
- ex/1+ex
- e^x/1+e^x
- e^x divide by (1+e^x)
-
Similar expressions
- e^x/(1-e^x)
Graphing y = e^x/(1+e^x)
The solution
You have entered
[src]
x
E
f(x) = ------
x
1 + E
$$f{\left(x \right)} = \frac{e^{x}}{e^{x} + 1}$$
f = E^x/(E^x + 1)
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:
$$\frac{e^{x}}{e^{x} + 1} = 0$$
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:
$$\frac{e^{x}}{e^{x} + 1} = 0$$
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
$$\frac{e^{x}}{e^{x} + 1} - \frac{e^{2 x}}{\left(e^{x} + 1\right)^{2}} = 0$$
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
$$\frac{e^{x}}{e^{x} + 1} - \frac{e^{2 x}}{\left(e^{x} + 1\right)^{2}} = 0$$
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
$$\frac{\left(- \frac{\left(1 - \frac{2 e^{x}}{e^{x} + 1}\right) e^{x}}{e^{x} + 1} + 1 - \frac{2 e^{x}}{e^{x} + 1}\right) e^{x}}{e^{x} + 1} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 0$$
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
$$\left(-\infty, 0\right]$$
Convex at the intervals
$$\left[0, \infty\right)$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{\left(- \frac{\left(1 - \frac{2 e^{x}}{e^{x} + 1}\right) e^{x}}{e^{x} + 1} + 1 - \frac{2 e^{x}}{e^{x} + 1}\right) e^{x}}{e^{x} + 1} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 0$$
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
$$\left(-\infty, 0\right]$$
Convex at the intervals
$$\left[0, \infty\right)$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{e^{x}}{e^{x} + 1}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{e^{x}}{e^{x} + 1}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
$$\lim_{x \to -\infty}\left(\frac{e^{x}}{e^{x} + 1}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{e^{x}}{e^{x} + 1}\right) = 1$$
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 E^x/(1 + E^x), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{e^{x}}{x \left(e^{x} + 1\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{e^{x}}{x \left(e^{x} + 1\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{e^{x}}{x \left(e^{x} + 1\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{e^{x}}{x \left(e^{x} + 1\right)}\right) = 0$$
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:
$$\frac{e^{x}}{e^{x} + 1} = \frac{e^{- x}}{1 + e^{- x}}$$
- No
$$\frac{e^{x}}{e^{x} + 1} = - \frac{e^{- x}}{1 + e^{- x}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{e^{x}}{e^{x} + 1} = \frac{e^{- x}}{1 + e^{- x}}$$
- No
$$\frac{e^{x}}{e^{x} + 1} = - \frac{e^{- x}}{1 + e^{- x}}$$
- No
so, the function
not is
neither even, nor odd