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Plotting a function graph/ tan(1/x)/tan(1/(1+x))

Graphing y = tan(1/x)/tan(1/(1+x))

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
            /1\  
         tan|-|  
            \x/  
f(x) = ----------
          /  1  \
       tan|-----|
          \1 + x/
f(x)=tan⁡(1x)tan⁡(1x+1)f{\left(x \right)} = \frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{x + 1} \right)}} 
f = tan(1/x)/tan(1/(x + 1))
The graph of the function
02468-8-6-4-2-1010-100100
The domain of the function
The points at which the function is not precisely defined:
x1=−1x_{1} = -1
x2=0x_{2} = 0
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:
tan⁡(1x)tan⁡(1x+1)=0\frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{x + 1} \right)}} = 0
Solve this equation
The points of intersection with the axis X:

Numerical solution
x1=−1.63661977236758x_{1} = -1.63661977236758
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to tan(1/x)/tan(1/(1 + x)).
tan⁡(10)tan⁡(1−1)\frac{\tan{\left(\frac{1}{0} \right)}}{\tan{\left(1^{-1} \right)}}
The result:
f(0)=NaNf{\left(0 \right)} = \text{NaN}
- the solutions of the equation d'not exist
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
(tan⁡2(1x+1)+1)tan⁡(1x)(x+1)2tan⁡2(1x+1)−tan⁡2(1x)+1x2tan⁡(1x+1)=0\frac{\left(\tan^{2}{\left(\frac{1}{x + 1} \right)} + 1\right) \tan{\left(\frac{1}{x} \right)}}{\left(x + 1\right)^{2} \tan^{2}{\left(\frac{1}{x + 1} \right)}} - \frac{\tan^{2}{\left(\frac{1}{x} \right)} + 1}{x^{2} \tan{\left(\frac{1}{x + 1} \right)}} = 0
Solve this equation
Solutions are not found,
function may have no extrema
Vertical asymptotes
Have:
x1=−1x_{1} = -1
x2=0x_{2} = 0
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(tan⁡(1x)tan⁡(1x+1))=1\lim_{x \to -\infty}\left(\frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{x + 1} \right)}}\right) = 1
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=1y = 1
lim⁡x→∞(tan⁡(1x)tan⁡(1x+1))=1\lim_{x \to \infty}\left(\frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{x + 1} \right)}}\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 tan(1/x)/tan(1/(1 + x)), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(tan⁡(1x)xtan⁡(1x+1))=0\lim_{x \to -\infty}\left(\frac{\tan{\left(\frac{1}{x} \right)}}{x \tan{\left(\frac{1}{x + 1} \right)}}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(tan⁡(1x)xtan⁡(1x+1))=0\lim_{x \to \infty}\left(\frac{\tan{\left(\frac{1}{x} \right)}}{x \tan{\left(\frac{1}{x + 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:
tan⁡(1x)tan⁡(1x+1)=−tan⁡(1x)tan⁡(11−x)\frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{x + 1} \right)}} = - \frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{1 - x} \right)}}
- No
tan⁡(1x)tan⁡(1x+1)=tan⁡(1x)tan⁡(11−x)\frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{x + 1} \right)}} = \frac{\tan{\left(\frac{1}{x} \right)}}{\tan{\left(\frac{1}{1 - x} \right)}}
- No
so, the function
not is
neither even, nor odd
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