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

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

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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        2*x + 1 
f(x) = ---------
       x*(x + 1)
f(x)=2x+1x(x+1)f{\left(x \right)} = \frac{2 x + 1}{x \left(x + 1\right)} 
f = (2*x + 1)/((x*(x + 1)))
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=−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:
2x+1x(x+1)=0\frac{2 x + 1}{x \left(x + 1\right)} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=−12x_{1} = - \frac{1}{2}
Numerical solution
x1=−0.5x_{1} = -0.5
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (2*x + 1)/((x*(x + 1))).
0⋅2+10\frac{0 \cdot 2 + 1}{0}
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
2x(x+1)+(−2x−1)(2x+1)x2(x+1)2=0\frac{2}{x \left(x + 1\right)} + \frac{\left(- 2 x - 1\right) \left(2 x + 1\right)}{x^{2} \left(x + 1\right)^{2}} = 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
(2x+1)((2x+1)(1x+1+1x)−6+2x+1x+1+2x+1x)x2(x+1)2=0\frac{\left(2 x + 1\right) \left(\left(2 x + 1\right) \left(\frac{1}{x + 1} + \frac{1}{x}\right) - 6 + \frac{2 x + 1}{x + 1} + \frac{2 x + 1}{x}\right)}{x^{2} \left(x + 1\right)^{2}} = 0
Solve this equation
The roots of this equation
x1=−12x_{1} = - \frac{1}{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=−1x_{1} = -1
x2=0x_{2} = 0

lim⁡x→−1−((2x+1)((2x+1)(1x+1+1x)−6+2x+1x+1+2x+1x)x2(x+1)2)=−∞\lim_{x \to -1^-}\left(\frac{\left(2 x + 1\right) \left(\left(2 x + 1\right) \left(\frac{1}{x + 1} + \frac{1}{x}\right) - 6 + \frac{2 x + 1}{x + 1} + \frac{2 x + 1}{x}\right)}{x^{2} \left(x + 1\right)^{2}}\right) = -\infty
lim⁡x→−1+((2x+1)((2x+1)(1x+1+1x)−6+2x+1x+1+2x+1x)x2(x+1)2)=∞\lim_{x \to -1^+}\left(\frac{\left(2 x + 1\right) \left(\left(2 x + 1\right) \left(\frac{1}{x + 1} + \frac{1}{x}\right) - 6 + \frac{2 x + 1}{x + 1} + \frac{2 x + 1}{x}\right)}{x^{2} \left(x + 1\right)^{2}}\right) = \infty
- the limits are not equal, so
x1=−1x_{1} = -1
- is an inflection point
lim⁡x→0−((2x+1)((2x+1)(1x+1+1x)−6+2x+1x+1+2x+1x)x2(x+1)2)=−∞\lim_{x \to 0^-}\left(\frac{\left(2 x + 1\right) \left(\left(2 x + 1\right) \left(\frac{1}{x + 1} + \frac{1}{x}\right) - 6 + \frac{2 x + 1}{x + 1} + \frac{2 x + 1}{x}\right)}{x^{2} \left(x + 1\right)^{2}}\right) = -\infty
lim⁡x→0+((2x+1)((2x+1)(1x+1+1x)−6+2x+1x+1+2x+1x)x2(x+1)2)=∞\lim_{x \to 0^+}\left(\frac{\left(2 x + 1\right) \left(\left(2 x + 1\right) \left(\frac{1}{x + 1} + \frac{1}{x}\right) - 6 + \frac{2 x + 1}{x + 1} + \frac{2 x + 1}{x}\right)}{x^{2} \left(x + 1\right)^{2}}\right) = \infty
- the limits are not equal, so
x2=0x_{2} = 0
- is an inflection 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
(−∞,−12]\left(-\infty, - \frac{1}{2}\right]
Convex at the intervals
[−12,∞)\left[- \frac{1}{2}, \infty\right)
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→−∞(2x+1x(x+1))=0\lim_{x \to -\infty}\left(\frac{2 x + 1}{x \left(x + 1\right)}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞(2x+1x(x+1))=0\lim_{x \to \infty}\left(\frac{2 x + 1}{x \left(x + 1\right)}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=0y = 0
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (2*x + 1)/((x*(x + 1))), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(1x(x+1)(2x+1)x)=0\lim_{x \to -\infty}\left(\frac{\frac{1}{x \left(x + 1\right)} \left(2 x + 1\right)}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(1x(x+1)(2x+1)x)=0\lim_{x \to \infty}\left(\frac{\frac{1}{x \left(x + 1\right)} \left(2 x + 1\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:
2x+1x(x+1)=−1−2xx(1−x)\frac{2 x + 1}{x \left(x + 1\right)} = - \frac{1 - 2 x}{x \left(1 - x\right)}
- No
2x+1x(x+1)=1−2xx(1−x)\frac{2 x + 1}{x \left(x + 1\right)} = \frac{1 - 2 x}{x \left(1 - x\right)}
- No
so, the function
not is
neither even, nor odd
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