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