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Plotting a function graph/ |x+2/x+3|

Graphing y = |x+2/x+3|

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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       |    2    |
f(x) = |x + - + 3|
       |    x    |
f(x)=∣(x+2x)+3∣f{\left(x \right)} = \left|{\left(x + \frac{2}{x}\right) + 3}\right| 
f = |x + 2/x + 3|
The graph of the function
02468-8-6-4-2-1010050
The domain of the function
The points at which the function is not precisely defined:
x1=0x_{1} = 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:
∣(x+2x)+3∣=0\left|{\left(x + \frac{2}{x}\right) + 3}\right| = 0
Solve this equation
The points of intersection with the axis X:

Numerical solution
x1=−2x_{1} = -2
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to |x + 2/x + 3|.
∣20+3∣\left|{\frac{2}{0} + 3}\right|
The result:
f(0)=∞f{\left(0 \right)} = \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
(1−2x2)sign⁡(x+3+2x)=0\left(1 - \frac{2}{x^{2}}\right) \operatorname{sign}{\left(x + 3 + \frac{2}{x} \right)} = 0
Solve this equation
The roots of this equation
x1=−2x_{1} = - \sqrt{2}
x2=2x_{2} = \sqrt{2}
The values of the extrema at the points:
    ___          ___ 
(-\/ 2, 3 - 2*\/ 2 )

   ___          ___ 
(\/ 2, 3 + 2*\/ 2 )


Intervals of increase and decrease of the function:
Let's find intervals where the function increases and decreases, as well as minima and maxima of the function, for this let's look how the function behaves itself in the extremas and at the slightest deviation from:
Minima of the function at points:
x1=2x_{1} = \sqrt{2}
Maxima of the function at points:
x1=−2x_{1} = - \sqrt{2}
Decreasing at intervals
(−∞,−2]∪[2,∞)\left(-\infty, - \sqrt{2}\right] \cup \left[\sqrt{2}, \infty\right)
Increasing at intervals
[−2,2]\left[- \sqrt{2}, \sqrt{2}\right]
Vertical asymptotes
Have:
x1=0x_{1} = 0
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞∣(x+2x)+3∣=∞\lim_{x \to -\infty} \left|{\left(x + \frac{2}{x}\right) + 3}\right| = \infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞∣(x+2x)+3∣=∞\lim_{x \to \infty} \left|{\left(x + \frac{2}{x}\right) + 3}\right| = \infty
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of |x + 2/x + 3|, divided by x at x->+oo and x ->-oo
lim⁡x→−∞(∣(x+2x)+3∣x)=−1\lim_{x \to -\infty}\left(\frac{\left|{\left(x + \frac{2}{x}\right) + 3}\right|}{x}\right) = -1
Let's take the limit
so,
inclined asymptote equation on the left:
y=−xy = - x
lim⁡x→∞(∣(x+2x)+3∣x)=1\lim_{x \to \infty}\left(\frac{\left|{\left(x + \frac{2}{x}\right) + 3}\right|}{x}\right) = 1
Let's take the limit
so,
inclined asymptote equation on the right:
y=xy = x
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:
∣(x+2x)+3∣=∣x−3+2x∣\left|{\left(x + \frac{2}{x}\right) + 3}\right| = \left|{x - 3 + \frac{2}{x}}\right|
- No
∣(x+2x)+3∣=−∣x−3+2x∣\left|{\left(x + \frac{2}{x}\right) + 3}\right| = - \left|{x - 3 + \frac{2}{x}}\right|
- No
so, the function
not is
neither even, nor odd
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