Mister Exam
Graphing y = (ln(x)-1)/(ln(x)+1)
The solution
You have entered
[src]
log(x) - 1
f(x) = ----------
log(x) + 1
$$f{\left(x \right)} = \frac{\log{\left(x \right)} - 1}{\log{\left(x \right)} + 1}$$
f = (log(x) - 1)/(log(x) + 1)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 0.367879441171442$$
$$x_{1} = 0.367879441171442$$
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:
$$\frac{\log{\left(x \right)} - 1}{\log{\left(x \right)} + 1} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = e$$
Numerical solution
$$x_{1} = 2.71828182845905$$
so we need to solve the equation:
$$\frac{\log{\left(x \right)} - 1}{\log{\left(x \right)} + 1} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = e$$
Numerical solution
$$x_{1} = 2.71828182845905$$
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
$$- \frac{\log{\left(x \right)} - 1}{x \left(\log{\left(x \right)} + 1\right)^{2}} + \frac{1}{x \left(\log{\left(x \right)} + 1\right)} = 0$$
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
$$- \frac{\log{\left(x \right)} - 1}{x \left(\log{\left(x \right)} + 1\right)^{2}} + \frac{1}{x \left(\log{\left(x \right)} + 1\right)} = 0$$
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
$$\frac{\frac{\left(1 + \frac{2}{\log{\left(x \right)} + 1}\right) \left(\log{\left(x \right)} - 1\right)}{\log{\left(x \right)} + 1} - 1 - \frac{2}{\log{\left(x \right)} + 1}}{x^{2} \left(\log{\left(x \right)} + 1\right)} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = e^{-3}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0.367879441171442$$
$$\lim_{x \to 0.367879441171442^-}\left(\frac{\frac{\left(1 + \frac{2}{\log{\left(x \right)} + 1}\right) \left(\log{\left(x \right)} - 1\right)}{\log{\left(x \right)} + 1} - 1 - \frac{2}{\log{\left(x \right)} + 1}}{x^{2} \left(\log{\left(x \right)} + 1\right)}\right) = \infty$$
$$\lim_{x \to 0.367879441171442^+}\left(\frac{\frac{\left(1 + \frac{2}{\log{\left(x \right)} + 1}\right) \left(\log{\left(x \right)} - 1\right)}{\log{\left(x \right)} + 1} - 1 - \frac{2}{\log{\left(x \right)} + 1}}{x^{2} \left(\log{\left(x \right)} + 1\right)}\right) = -\infty$$
- the limits are not equal, so
$$x_{1} = 0.367879441171442$$
- is an inflection point
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
$$\left[e^{-3}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, e^{-3}\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{\frac{\left(1 + \frac{2}{\log{\left(x \right)} + 1}\right) \left(\log{\left(x \right)} - 1\right)}{\log{\left(x \right)} + 1} - 1 - \frac{2}{\log{\left(x \right)} + 1}}{x^{2} \left(\log{\left(x \right)} + 1\right)} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = e^{-3}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0.367879441171442$$
$$\lim_{x \to 0.367879441171442^-}\left(\frac{\frac{\left(1 + \frac{2}{\log{\left(x \right)} + 1}\right) \left(\log{\left(x \right)} - 1\right)}{\log{\left(x \right)} + 1} - 1 - \frac{2}{\log{\left(x \right)} + 1}}{x^{2} \left(\log{\left(x \right)} + 1\right)}\right) = \infty$$
$$\lim_{x \to 0.367879441171442^+}\left(\frac{\frac{\left(1 + \frac{2}{\log{\left(x \right)} + 1}\right) \left(\log{\left(x \right)} - 1\right)}{\log{\left(x \right)} + 1} - 1 - \frac{2}{\log{\left(x \right)} + 1}}{x^{2} \left(\log{\left(x \right)} + 1\right)}\right) = -\infty$$
- the limits are not equal, so
$$x_{1} = 0.367879441171442$$
- is an inflection point
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
$$\left[e^{-3}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, e^{-3}\right]$$
Vertical asymptotes
Have:
$$x_{1} = 0.367879441171442$$
$$x_{1} = 0.367879441171442$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{\log{\left(x \right)} - 1}{\log{\left(x \right)} + 1}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{\log{\left(x \right)} - 1}{\log{\left(x \right)} + 1}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
$$\lim_{x \to -\infty}\left(\frac{\log{\left(x \right)} - 1}{\log{\left(x \right)} + 1}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{\log{\left(x \right)} - 1}{\log{\left(x \right)} + 1}\right) = 1$$
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 (log(x) - 1)/(log(x) + 1), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\log{\left(x \right)} - 1}{x \left(\log{\left(x \right)} + 1\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\log{\left(x \right)} - 1}{x \left(\log{\left(x \right)} + 1\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\log{\left(x \right)} - 1}{x \left(\log{\left(x \right)} + 1\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\log{\left(x \right)} - 1}{x \left(\log{\left(x \right)} + 1\right)}\right) = 0$$
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:
$$\frac{\log{\left(x \right)} - 1}{\log{\left(x \right)} + 1} = \frac{\log{\left(- x \right)} - 1}{\log{\left(- x \right)} + 1}$$
- No
$$\frac{\log{\left(x \right)} - 1}{\log{\left(x \right)} + 1} = - \frac{\log{\left(- x \right)} - 1}{\log{\left(- x \right)} + 1}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{\log{\left(x \right)} - 1}{\log{\left(x \right)} + 1} = \frac{\log{\left(- x \right)} - 1}{\log{\left(- x \right)} + 1}$$
- No
$$\frac{\log{\left(x \right)} - 1}{\log{\left(x \right)} + 1} = - \frac{\log{\left(- x \right)} - 1}{\log{\left(- x \right)} + 1}$$
- No
so, the function
not is
neither even, nor odd