Mister Exam
Graphing y = sign((2-x)\(x+1))
The solution
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/2 - x\
f(x) = sign|-----|
\x + 1/
$$f{\left(x \right)} = \operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)}$$
f = sign((2 - x)/(x + 1))
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -1$$
$$x_{1} = -1$$
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:
$$\operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)} = 0$$
Solve this equation
The points of intersection with the axis X:
Numerical solution
$$x_{1} = 2$$
so we need to solve the equation:
$$\operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)} = 0$$
Solve this equation
The points of intersection with the axis X:
Numerical solution
$$x_{1} = 2$$
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
$$\frac{d}{d x} \operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(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
$$\frac{d}{d x} \operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
Vertical asymptotes
Have:
$$x_{1} = -1$$
$$x_{1} = -1$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty} \operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)} = -1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = -1$$
$$\lim_{x \to -\infty} \operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty} \operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)} = -1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = -1$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of sign((2 - x)/(x + 1)), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)}}{x}\right) = 0$$
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:
$$\operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)} = \operatorname{sign}{\left(\frac{x + 2}{1 - x} \right)}$$
- No
$$\operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)} = - \operatorname{sign}{\left(\frac{x + 2}{1 - x} \right)}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)} = \operatorname{sign}{\left(\frac{x + 2}{1 - x} \right)}$$
- No
$$\operatorname{sign}{\left(\frac{2 - x}{x + 1} \right)} = - \operatorname{sign}{\left(\frac{x + 2}{1 - x} \right)}$$
- No
so, the function
not is
neither even, nor odd