Mister Exam
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-
How to use it?
- Graphing y =:
- ||x-4|-2|
- |x-3|-|x+3|
- x^3+x^2-8x+1
- x^3-9x^2+24x-16
- Derivative of:
-
1/(1+e^(-x))
- Integral of d{x}:
-
1/(1+e^(-x))
- Limit of the function:
-
1/(1+e^(-x))
-
Identical expressions
- one /(one +e^(-x))
- 1 divide by (1 plus e to the power of ( minus x))
- one divide by (one plus e to the power of ( minus x))
- 1/(1+e(-x))
- 1/1+e-x
- 1/1+e^-x
- 1 divide by (1+e^(-x))
-
Similar expressions
- 1/(1-e^(-x))
- 1/(1+e^(x))
Graphing y = 1/(1+e^(-x))
The solution
You have entered
[src]
1
f(x) = -------
-x
1 + E
$$f{\left(x \right)} = \frac{1}{1 + e^{- x}}$$
f = 1/(1 + E^(-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:
$$\frac{1}{1 + e^{- x}} = 0$$
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:
$$\frac{1}{1 + e^{- x}} = 0$$
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
$$\frac{e^{- x}}{\left(1 + e^{- x}\right)^{2}} = 0$$
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
$$\frac{e^{- x}}{\left(1 + e^{- x}\right)^{2}} = 0$$
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
$$\frac{\left(-1 + \frac{2 e^{- x}}{1 + e^{- x}}\right) e^{- x}}{\left(1 + e^{- x}\right)^{2}} = 0$$
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
$$\frac{\left(-1 + \frac{2 e^{- x}}{1 + e^{- x}}\right) e^{- x}}{\left(1 + e^{- x}\right)^{2}} = 0$$
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
$$\lim_{x \to -\infty} \frac{1}{1 + e^{- x}} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty} \frac{1}{1 + e^{- x}} = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
$$\lim_{x \to -\infty} \frac{1}{1 + e^{- x}} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty} \frac{1}{1 + e^{- x}} = 1$$
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 1/(1 + E^(-x)), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{1}{x \left(1 + e^{- x}\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{1}{x \left(1 + e^{- x}\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{1}{x \left(1 + e^{- x}\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{1}{x \left(1 + e^{- x}\right)}\right) = 0$$
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:
$$\frac{1}{1 + e^{- x}} = \frac{1}{e^{x} + 1}$$
- No
$$\frac{1}{1 + e^{- x}} = - \frac{1}{e^{x} + 1}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{1}{1 + e^{- x}} = \frac{1}{e^{x} + 1}$$
- No
$$\frac{1}{1 + e^{- x}} = - \frac{1}{e^{x} + 1}$$
- No
so, the function
not is
neither even, nor odd