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  • Graphing y =:
  • x(2-x)^2
  • x√2-x
  • -x²+6x-5
  • x^2-8x
  • Identical expressions

  • one /(ln(x)/(x- three))
  • 1 divide by (ln(x) divide by (x minus 3))
  • one divide by (ln(x) divide by (x minus three))
  • 1/lnx/x-3
  • 1 divide by (ln(x) divide by (x-3))
  • Similar expressions

  • 1/(ln(x)/(x+3))

Graphing y = 1/(ln(x)/(x-3))

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src]
          1    
f(x) = --------
       /log(x)\
       |------|
       \x - 3 /
$$f{\left(x \right)} = \frac{1}{\frac{1}{x - 3} \log{\left(x \right)}}$$
f = 1/(log(x)/(x - 3))
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 1$$
$$x_{2} = 3$$
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:
$$\frac{1}{\frac{1}{x - 3} \log{\left(x \right)}} = 0$$
Solve this equation
The points of intersection with the axis X:

Analytical solution
$$x_{1} = 3$$
Numerical solution
$$x_{1} = 3$$
$$x_{2} = 0$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 1/(log(x)/(x - 3)).
$$\frac{1}{\frac{1}{-3} \log{\left(0 \right)}}$$
The result:
$$f{\left(0 \right)} = 0$$
The point:
(0, 0)
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
$$\frac{\frac{x - 3}{\log{\left(x \right)}} \left(x - 3\right) \left(\frac{\log{\left(x \right)}}{\left(x - 3\right)^{2}} - \frac{1}{x \left(x - 3\right)}\right)}{\log{\left(x \right)}} = 0$$
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
$$\frac{\left(1 - \frac{x - 3}{x \log{\left(x \right)}}\right) \left(\frac{\log{\left(x \right)}}{x - 3} - \frac{1}{x}\right) + \left(x - 3\right) \left(- \frac{2 \log{\left(x \right)}}{\left(x - 3\right)^{2}} + \frac{2}{x \left(x - 3\right)} + \frac{1}{x^{2}}\right) + \frac{\log{\left(x \right)}}{x - 3} - \frac{\left(x - 3\right) \left(\frac{\log{\left(x \right)}}{x - 3} - \frac{1}{x}\right)}{x \log{\left(x \right)}} - \frac{1}{x}}{\log{\left(x \right)}^{2}} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
$$x_{1} = 1$$
$$x_{2} = 3$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \frac{1}{\frac{1}{x - 3} \log{\left(x \right)}} = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty} \frac{1}{\frac{1}{x - 3} \log{\left(x \right)}} = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 1/(log(x)/(x - 3)), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(x - 3\right) \frac{1}{\log{\left(x \right)}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\left(x - 3\right) \frac{1}{\log{\left(x \right)}}}{x}\right) = 0$$
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:
$$\frac{1}{\frac{1}{x - 3} \log{\left(x \right)}} = \frac{- x - 3}{\log{\left(- x \right)}}$$
- No
$$\frac{1}{\frac{1}{x - 3} \log{\left(x \right)}} = - \frac{- x - 3}{\log{\left(- x \right)}}$$
- No
so, the function
not is
neither even, nor odd