Mister Exam
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-
How to use it?
- Graphing y =:
- x^2
-
arcctg2x
- sin(5x+4)
- 100x
-
Identical expressions
- tanh^- one (x+ two)
- hyperbolic tangent of gent of to the power of minus 1(x plus 2)
- hyperbolic tangent of gent of to the power of minus one (x plus two)
- tanh-1(x+2)
- tanh-1x+2
- tanh^-1x+2
-
Similar expressions
- tanh^+1(x+2)
- tanh^-1(x-2)
Graphing y = tanh^-1(x+2)
The solution
You have entered
[src]
1
f(x) = -----------
tanh(x + 2)
$$f{\left(x \right)} = \frac{1}{\tanh{\left(x + 2 \right)}}$$
f = 1/tanh(x + 2)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -2$$
$$x_{1} = -2$$
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:
$$\frac{1}{\tanh{\left(x + 2 \right)}} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
so we need to solve the equation:
$$\frac{1}{\tanh{\left(x + 2 \right)}} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
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
$$\frac{\tanh^{2}{\left(x + 2 \right)} - 1}{\tanh^{2}{\left(x + 2 \right)}} = 0$$
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
$$\frac{\tanh^{2}{\left(x + 2 \right)} - 1}{\tanh^{2}{\left(x + 2 \right)}} = 0$$
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
$$\frac{2 \left(-1 + \frac{\tanh^{2}{\left(x + 2 \right)} - 1}{\tanh^{2}{\left(x + 2 \right)}}\right) \left(\tanh^{2}{\left(x + 2 \right)} - 1\right)}{\tanh{\left(x + 2 \right)}} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{2 \left(-1 + \frac{\tanh^{2}{\left(x + 2 \right)} - 1}{\tanh^{2}{\left(x + 2 \right)}}\right) \left(\tanh^{2}{\left(x + 2 \right)} - 1\right)}{\tanh{\left(x + 2 \right)}} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
$$x_{1} = -2$$
$$x_{1} = -2$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \frac{1}{\tanh{\left(x + 2 \right)}} = -1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = -1$$
$$\lim_{x \to \infty} \frac{1}{\tanh{\left(x + 2 \right)}} = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
$$\lim_{x \to -\infty} \frac{1}{\tanh{\left(x + 2 \right)}} = -1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = -1$$
$$\lim_{x \to \infty} \frac{1}{\tanh{\left(x + 2 \right)}} = 1$$
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 1/tanh(x + 2), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{1}{x \tanh{\left(x + 2 \right)}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{1}{x \tanh{\left(x + 2 \right)}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{1}{x \tanh{\left(x + 2 \right)}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{1}{x \tanh{\left(x + 2 \right)}}\right) = 0$$
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:
$$\frac{1}{\tanh{\left(x + 2 \right)}} = - \frac{1}{\tanh{\left(x - 2 \right)}}$$
- No
$$\frac{1}{\tanh{\left(x + 2 \right)}} = \frac{1}{\tanh{\left(x - 2 \right)}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{1}{\tanh{\left(x + 2 \right)}} = - \frac{1}{\tanh{\left(x - 2 \right)}}$$
- No
$$\frac{1}{\tanh{\left(x + 2 \right)}} = \frac{1}{\tanh{\left(x - 2 \right)}}$$
- No
so, the function
not is
neither even, nor odd
The graph