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  • How to use it?

  • Graphing y =:
  • x/(4-x^2)
  • x^4-6x^2+1
  • x^4-4x^3+4
  • -x^4+4x^2-5
  • Limit of the function:
  • 2^(x/(9-x^2)) 2^(x/(9-x^2))

  • Identical expressions

  • two ^(x/(nine -x^ two))
  • 2 to the power of (x divide by (9 minus x squared ))
  • two to the power of (x divide by (nine minus x to the power of two))
  • 2(x/(9-x2))
  • 2x/9-x2
  • 2^(x/(9-x²))
  • 2 to the power of (x/(9-x to the power of 2))
  • 2^x/9-x^2
  • 2^(x divide by (9-x^2))

  • Similar expressions

  • 2^(x/(9+x^2))
Plotting a function graph/ 2^(x/(9-x^2))

Graphing y = 2^(x/(9-x^2))

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
          x   
        ------
             2
        9 - x 
f(x) = 2
$$f{\left(x \right)} = 2^{\frac{x}{9 - x^{2}}}$$ 
f = 2^(x/(9 - x^2))
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -3$$
$$x_{2} = 3$$
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:
$$2^{\frac{x}{9 - x^{2}}} = 0$$
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 2^(x/(9 - x^2)).
$$2^{\frac{0}{9 - 0^{2}}}$$
The result:
$$f{\left(0 \right)} = 1$$
The point:
(0, 1)
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
$$2^{\frac{x}{9 - x^{2}}} \left(\frac{2 x^{2}}{\left(9 - x^{2}\right)^{2}} + \frac{1}{9 - x^{2}}\right) \log{\left(2 \right)} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
Vertical asymptotes
Have:
$$x_{1} = -3$$
$$x_{2} = 3$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} 2^{\frac{x}{9 - x^{2}}} = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty} 2^{\frac{x}{9 - x^{2}}} = 1$$
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/(9 - x^2)), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{2^{\frac{x}{9 - x^{2}}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{2^{\frac{x}{9 - x^{2}}}}{x}\right) = 0$$
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:
$$2^{\frac{x}{9 - x^{2}}} = 2^{- \frac{x}{9 - x^{2}}}$$
- No
$$2^{\frac{x}{9 - x^{2}}} = - 2^{- \frac{x}{9 - x^{2}}}$$
- No
so, the function
not is
neither even, nor odd
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