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Plotting a function graph/ 1-ln(x/(x+3))

Graphing y = 1-ln(x/(x+3))

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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              /  x  \
f(x) = 1 - log|-----|
              \x + 3/
f(x)=1−log⁡(xx+3)f{\left(x \right)} = 1 - \log{\left(\frac{x}{x + 3} \right)} 
f = 1 - log(x/(x + 3))
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=−3x_{1} = -3
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:
1−log⁡(xx+3)=01 - \log{\left(\frac{x}{x + 3} \right)} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=3e1−ex_{1} = \frac{3 e}{1 - e}
Numerical solution
x1=−4.74593012060798x_{1} = -4.74593012060798
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 1 - log(x/(x + 3)).
1−log⁡(03)1 - \log{\left(\frac{0}{3} \right)}
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+3)(−x(x+3)2+1x+3)x=0- \frac{\left(x + 3\right) \left(- \frac{x}{\left(x + 3\right)^{2}} + \frac{1}{x + 3}\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+3−1)(−1x+3−1x)x=0\frac{\left(\frac{x}{x + 3} - 1\right) \left(- \frac{1}{x + 3} - \frac{1}{x}\right)}{x} = 0
Solve this equation
The roots of this equation
x1=−32x_{1} = - \frac{3}{2}
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=−3x_{1} = -3

lim⁡x→−3−((xx+3−1)(−1x+3−1x)x)=−∞\lim_{x \to -3^-}\left(\frac{\left(\frac{x}{x + 3} - 1\right) \left(- \frac{1}{x + 3} - \frac{1}{x}\right)}{x}\right) = -\infty
lim⁡x→−3+((xx+3−1)(−1x+3−1x)x)=−∞\lim_{x \to -3^+}\left(\frac{\left(\frac{x}{x + 3} - 1\right) \left(- \frac{1}{x + 3} - \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
[−32,∞)\left[- \frac{3}{2}, \infty\right)
Convex at the intervals
(−∞,−32]\left(-\infty, - \frac{3}{2}\right]
Vertical asymptotes
Have:
x1=−3x_{1} = -3
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(1−log⁡(xx+3))=1\lim_{x \to -\infty}\left(1 - \log{\left(\frac{x}{x + 3} \right)}\right) = 1
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=1y = 1
lim⁡x→∞(1−log⁡(xx+3))=1\lim_{x \to \infty}\left(1 - \log{\left(\frac{x}{x + 3} \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 1 - log(x/(x + 3)), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(1−log⁡(xx+3)x)=0\lim_{x \to -\infty}\left(\frac{1 - \log{\left(\frac{x}{x + 3} \right)}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(1−log⁡(xx+3)x)=0\lim_{x \to \infty}\left(\frac{1 - \log{\left(\frac{x}{x + 3} \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:
1−log⁡(xx+3)=1−log⁡(−x3−x)1 - \log{\left(\frac{x}{x + 3} \right)} = 1 - \log{\left(- \frac{x}{3 - x} \right)}
- No
1−log⁡(xx+3)=log⁡(−x3−x)−11 - \log{\left(\frac{x}{x + 3} \right)} = \log{\left(- \frac{x}{3 - x} \right)} - 1
- No
so, the function
not is
neither even, nor odd
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