Mister Exam
Graphing y = (2^x+2^(-x))/(2^x+2^(-x))
The solution
You have entered
[src]
x -x
2 + 2
f(x) = --------
x -x
2 + 2
$$f{\left(x \right)} = \frac{2^{x} + 2^{- x}}{2^{x} + 2^{- x}}$$
f = (2^x + 2^(-x))/(2^x + 2^(-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:
$$\frac{2^{x} + 2^{- x}}{2^{x} + 2^{- x}} = 0$$
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:
$$\frac{2^{x} + 2^{- x}}{2^{x} + 2^{- x}} = 0$$
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
$$\frac{- 2^{x} \log{\left(2 \right)} + 2^{- x} \log{\left(2 \right)}}{2^{x} + 2^{- x}} + \frac{2^{x} \log{\left(2 \right)} - 2^{- x} \log{\left(2 \right)}}{2^{x} + 2^{- x}} = 0$$
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
$$\frac{- 2^{x} \log{\left(2 \right)} + 2^{- x} \log{\left(2 \right)}}{2^{x} + 2^{- x}} + \frac{2^{x} \log{\left(2 \right)} - 2^{- x} \log{\left(2 \right)}}{2^{x} + 2^{- x}} = 0$$
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
$$0 = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$0 = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{2^{x} + 2^{- x}}{2^{x} + 2^{- x}}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{2^{x} + 2^{- x}}{2^{x} + 2^{- x}}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
$$\lim_{x \to -\infty}\left(\frac{2^{x} + 2^{- x}}{2^{x} + 2^{- x}}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{2^{x} + 2^{- x}}{2^{x} + 2^{- x}}\right) = 1$$
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 (2^x + 2^(-x))/(2^x + 2^(-x)), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty} \frac{1}{x} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty} \frac{1}{x} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty} \frac{1}{x} = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty} \frac{1}{x} = 0$$
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:
$$\frac{2^{x} + 2^{- x}}{2^{x} + 2^{- x}} = \frac{2^{x} + 2^{- x}}{2^{x} + 2^{- x}}$$
- Yes
$$\frac{2^{x} + 2^{- x}}{2^{x} + 2^{- x}} = - \frac{2^{x} + 2^{- x}}{2^{x} + 2^{- x}}$$
- No
so, the function
is
even
So, check:
$$\frac{2^{x} + 2^{- x}}{2^{x} + 2^{- x}} = \frac{2^{x} + 2^{- x}}{2^{x} + 2^{- x}}$$
- Yes
$$\frac{2^{x} + 2^{- x}}{2^{x} + 2^{- x}} = - \frac{2^{x} + 2^{- x}}{2^{x} + 2^{- x}}$$
- No
so, the function
is
even