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3/(2^x-1)
  • How to use it?

  • Graphing y =:
  • x/(4-x^2)
  • -x^4+x^2+5
  • x^4-6x^2+1
  • x^4-4x^3+4
  • Identical expressions

  • three /(two ^x- one)
  • 3 divide by (2 to the power of x minus 1)
  • three divide by (two to the power of x minus one)
  • 3/(2x-1)
  • 3/2x-1
  • 3/2^x-1
  • 3 divide by (2^x-1)
  • Similar expressions

  • 3/(2^x+1)

Graphing y = 3/(2^x-1)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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         3   
f(x) = ------
        x    
       2  - 1
$$f{\left(x \right)} = \frac{3}{2^{x} - 1}$$
f = 3/(2^x - 1*1)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 0$$
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:
$$\frac{3}{2^{x} - 1} = 0$$
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 3/(2^x - 1*1).
$$\frac{3}{\left(-1\right) 1 + 2^{0}}$$
The result:
$$f{\left(0 \right)} = \tilde{\infty}$$
sof doesn't intersect Y
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
$$- \frac{3 \cdot 2^{x} \log{\left(2 \right)}}{\left(2^{x} - 1\right)^{2}} = 0$$
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
$$\frac{3 \cdot 2^{x} \left(\frac{2 \cdot 2^{x}}{2^{x} - 1} - 1\right) \log{\left(2 \right)}^{2}}{\left(2^{x} - 1\right)^{2}} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
$$x_{1} = 0$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{3}{2^{x} - 1}\right) = -3$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = -3$$
$$\lim_{x \to \infty}\left(\frac{3}{2^{x} - 1}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 3/(2^x - 1*1), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{3}{x \left(2^{x} - 1\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{3}{x \left(2^{x} - 1\right)}\right) = 0$$
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:
$$\frac{3}{2^{x} - 1} = \frac{3}{-1 + 2^{- x}}$$
- No
$$\frac{3}{2^{x} - 1} = - \frac{3}{-1 + 2^{- x}}$$
- No
so, the function
not is
neither even, nor odd
The graph
Graphing y = 3/(2^x-1)