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Plotting a function graph/ ln(x/(x+2))+1

Graphing y = ln(x/(x+2))+1

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
          /  x  \    
f(x) = log|-----| + 1
          \x + 2/
f(x)=log⁡(xx+2)+1f{\left(x \right)} = \log{\left(\frac{x}{x + 2} \right)} + 1 
f = log(x/(x + 2)) + 1
The graph of the function
02468-8-6-4-2-1010-1010
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:
log⁡(xx+2)+1=0\log{\left(\frac{x}{x + 2} \right)} + 1 = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=2−1+ex_{1} = \frac{2}{-1 + e}
Numerical solution
x1=1.16395341373865x_{1} = 1.16395341373865
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to log(x/(x + 2)) + 1.
log⁡(02)+1\log{\left(\frac{0}{2} \right)} + 1
The result:
f(0)=∞~f{\left(0 \right)} = \tilde{\infty}
sof doesn't intersect Y
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
(x+2)(−x(x+2)2+1x+2)x=0\frac{\left(x + 2\right) \left(- \frac{x}{\left(x + 2\right)^{2}} + \frac{1}{x + 2}\right)}{x} = 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
(xx+2−1)(1x+2+1x)x=0\frac{\left(\frac{x}{x + 2} - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x}\right)}{x} = 0
Solve this equation
The roots of this equation
x1=−1x_{1} = -1
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
x1=−2x_{1} = -2

lim⁡x→−2−((xx+2−1)(1x+2+1x)x)=∞\lim_{x \to -2^-}\left(\frac{\left(\frac{x}{x + 2} - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x}\right)}{x}\right) = \infty
lim⁡x→−2+((xx+2−1)(1x+2+1x)x)=∞\lim_{x \to -2^+}\left(\frac{\left(\frac{x}{x + 2} - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x}\right)}{x}\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
(−∞,−1]\left(-\infty, -1\right]
Convex at the intervals
[−1,∞)\left[-1, \infty\right)
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→−∞(log⁡(xx+2)+1)=1\lim_{x \to -\infty}\left(\log{\left(\frac{x}{x + 2} \right)} + 1\right) = 1
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=1y = 1
lim⁡x→∞(log⁡(xx+2)+1)=1\lim_{x \to \infty}\left(\log{\left(\frac{x}{x + 2} \right)} + 1\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 log(x/(x + 2)) + 1, divided by x at x->+oo and x ->-oo
lim⁡x→−∞(log⁡(xx+2)+1x)=0\lim_{x \to -\infty}\left(\frac{\log{\left(\frac{x}{x + 2} \right)} + 1}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(log⁡(xx+2)+1x)=0\lim_{x \to \infty}\left(\frac{\log{\left(\frac{x}{x + 2} \right)} + 1}{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⁡(xx+2)+1=log⁡(−x2−x)+1\log{\left(\frac{x}{x + 2} \right)} + 1 = \log{\left(- \frac{x}{2 - x} \right)} + 1
- No
log⁡(xx+2)+1=−log⁡(−x2−x)−1\log{\left(\frac{x}{x + 2} \right)} + 1 = - \log{\left(- \frac{x}{2 - x} \right)} - 1
- No
so, the function
not is
neither even, nor odd
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