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Plotting a function graph/ |x+1|-|x-1|-x

Graphing y = |x+1|-|x-1|-x

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = |x + 1| - |x - 1| - x
f(x)=−x+(−∣x−1∣+∣x+1∣)f{\left(x \right)} = - x + \left(- \left|{x - 1}\right| + \left|{x + 1}\right|\right) 
f = -x - |x - 1| + |x + 1|
The graph of the function
02468-8-6-4-2-1010-2020
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+(−∣x−1∣+∣x+1∣)=0- x + \left(- \left|{x - 1}\right| + \left|{x + 1}\right|\right) = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=−2x_{1} = -2
x2=0x_{2} = 0
x3=2x_{3} = 2
Numerical solution
x1=−2x_{1} = -2
x2=0x_{2} = 0
x3=2x_{3} = 2
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to |x + 1| - |x - 1| - x.
−0+(−∣−1∣+∣1∣)- 0 + \left(- \left|{-1}\right| + \left|{1}\right|\right)
The result:
f(0)=0f{\left(0 \right)} = 0
The point:
(0, 0)
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
−sign⁡(x−1)+sign⁡(x+1)−1=0- \operatorname{sign}{\left(x - 1 \right)} + \operatorname{sign}{\left(x + 1 \right)} - 1 = 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
2(−δ(x−1)+δ(x+1))=02 \left(- \delta\left(x - 1\right) + \delta\left(x + 1\right)\right) = 0
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(−x+(−∣x−1∣+∣x+1∣))=∞\lim_{x \to -\infty}\left(- x + \left(- \left|{x - 1}\right| + \left|{x + 1}\right|\right)\right) = \infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞(−x+(−∣x−1∣+∣x+1∣))=−∞\lim_{x \to \infty}\left(- x + \left(- \left|{x - 1}\right| + \left|{x + 1}\right|\right)\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 + 1| - |x - 1| - x, divided by x at x->+oo and x ->-oo
lim⁡x→−∞(−x+(−∣x−1∣+∣x+1∣)x)=−1\lim_{x \to -\infty}\left(\frac{- x + \left(- \left|{x - 1}\right| + \left|{x + 1}\right|\right)}{x}\right) = -1
Let's take the limit
so,
inclined asymptote equation on the left:
y=−xy = - x
lim⁡x→∞(−x+(−∣x−1∣+∣x+1∣)x)=−1\lim_{x \to \infty}\left(\frac{- x + \left(- \left|{x - 1}\right| + \left|{x + 1}\right|\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+(−∣x−1∣+∣x+1∣)=x+∣x−1∣−∣x+1∣- x + \left(- \left|{x - 1}\right| + \left|{x + 1}\right|\right) = x + \left|{x - 1}\right| - \left|{x + 1}\right|
- No
−x+(−∣x−1∣+∣x+1∣)=−x−∣x−1∣+∣x+1∣- x + \left(- \left|{x - 1}\right| + \left|{x + 1}\right|\right) = - x - \left|{x - 1}\right| + \left|{x + 1}\right|
- No
so, the function
not is
neither even, nor odd
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