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tanh^-1(x+2)
Plotting a function graph/ tanh^-1(x+2)

Graphing y = tanh^-1(x+2)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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            1     
f(x) = -----------
       tanh(x + 2)
f(x)=1tanh⁡(x+2)f{\left(x \right)} = \frac{1}{\tanh{\left(x + 2 \right)}} 
f = 1/tanh(x + 2)
The graph of the function
02468-8-6-4-2-1010-5050
The domain of the function
The points at which the function is not precisely defined:
x1=−2x_{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:
1tanh⁡(x+2)=0\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
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 1/tanh(x + 2).
1tanh⁡(0+2)\frac{1}{\tanh{\left(0 + 2 \right)}}
The result:
f(0)=1tanh⁡(2)f{\left(0 \right)} = \frac{1}{\tanh{\left(2 \right)}}
The point:
(0, 1/tanh(2))
Extrema of the function
In order to find the extrema, we need to solve the equation
ddxf(x)=0\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:
ddxf(x)=\frac{d}{d x} f{\left(x \right)} =
the first derivative
tanh⁡2(x+2)−1tanh⁡2(x+2)=0\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
d2dx2f(x)=0\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:
d2dx2f(x)=\frac{d^{2}}{d x^{2}} f{\left(x \right)} =
the second derivative
2(−1+tanh⁡2(x+2)−1tanh⁡2(x+2))(tanh⁡2(x+2)−1)tanh⁡(x+2)=0\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:
x1=−2x_{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→−∞1tanh⁡(x+2)=−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=−1y = -1
lim⁡x→∞1tanh⁡(x+2)=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=1y = 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→−∞(1xtanh⁡(x+2))=0\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→∞(1xtanh⁡(x+2))=0\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:
1tanh⁡(x+2)=−1tanh⁡(x−2)\frac{1}{\tanh{\left(x + 2 \right)}} = - \frac{1}{\tanh{\left(x - 2 \right)}}
- No
1tanh⁡(x+2)=1tanh⁡(x−2)\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
Graphing y = tanh^-1(x+2)
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