Mister Exam

Graphing y = |x+3|-|x-1|+x+2

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = |x + 3| - |x - 1| + x + 2
$$f{\left(x \right)} = \left(x + \left(- \left|{x - 1}\right| + \left|{x + 3}\right|\right)\right) + 2$$
f = x - |x - 1| + |x + 3| + 2
The graph of the function
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:
$$\left(x + \left(- \left|{x - 1}\right| + \left|{x + 3}\right|\right)\right) + 2 = 0$$
Solve this equation
The points of intersection with the axis X:

Analytical solution
$$x_{1} = - \frac{4}{3}$$
Numerical solution
$$x_{1} = -1.33333333333333$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to |x + 3| - |x - 1| + x + 2.
$$\left(- \left|{-1}\right| + \left|{3}\right|\right) + 2$$
The result:
$$f{\left(0 \right)} = 4$$
The point:
(0, 4)
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} = $$
the first derivative
$$- \operatorname{sign}{\left(x - 1 \right)} + \operatorname{sign}{\left(x + 3 \right)} + 1 = 0$$
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
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative
$$2 \left(- \delta\left(x - 1\right) + \delta\left(x + 3\right)\right) = 0$$
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
$$\lim_{x \to -\infty}\left(\left(x + \left(- \left|{x - 1}\right| + \left|{x + 3}\right|\right)\right) + 2\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\left(x + \left(- \left|{x - 1}\right| + \left|{x + 3}\right|\right)\right) + 2\right) = \infty$$
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 + 3| - |x - 1| + x + 2, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(x + \left(- \left|{x - 1}\right| + \left|{x + 3}\right|\right)\right) + 2}{x}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x$$
$$\lim_{x \to \infty}\left(\frac{\left(x + \left(- \left|{x - 1}\right| + \left|{x + 3}\right|\right)\right) + 2}{x}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x$$
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:
$$\left(x + \left(- \left|{x - 1}\right| + \left|{x + 3}\right|\right)\right) + 2 = - x + \left|{x - 3}\right| - \left|{x + 1}\right| + 2$$
- No
$$\left(x + \left(- \left|{x - 1}\right| + \left|{x + 3}\right|\right)\right) + 2 = x - \left|{x - 3}\right| + \left|{x + 1}\right| - 2$$
- No
so, the function
not is
neither even, nor odd