Mister Exam
Other calculators
- Graph of implicit function
- Derivative Step by Step
- Integral Step by Step
- Graph of parametric function
- Graph in Polar Coordinates
- Surface defined parametrically
- Surface defined by equation
- Curve in 3D space defined parametrically
- Area between lines
- The intersection points of functions
-
How to use it?
- Graphing y =:
- (x+3)^3/(x+1)^2
- (x+2)/(x-1)
- x^2+6x+5
- x^2+5x-4
- Integral of d{x}:
- |-1+x|
- Limit of the function:
-
|-1+x|
-
Identical expressions
- |- one +x|
- module of minus 1 plus x|
- module of minus one plus x|
-
Similar expressions
- |+1+x|
- |-1-x|
Graphing y = |-1+x|
The solution
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:
$$\left|{x - 1}\right| = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 1$$
Numerical solution
$$x_{1} = 1$$
so we need to solve the equation:
$$\left|{x - 1}\right| = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 1$$
Numerical solution
$$x_{1} = 1$$
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
$$\operatorname{sign}{\left(x - 1 \right)} = 0$$
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
$$\operatorname{sign}{\left(x - 1 \right)} = 0$$
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
$$2 \delta\left(x - 1\right) = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$2 \delta\left(x - 1\right) = 0$$
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
$$\lim_{x \to -\infty} \left|{x - 1}\right| = \infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty} \left|{x - 1}\right| = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{x \to -\infty} \left|{x - 1}\right| = \infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty} \left|{x - 1}\right| = \infty$$
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 |-1 + x|, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\left|{x - 1}\right|}{x}\right) = -1$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = - x$$
$$\lim_{x \to \infty}\left(\frac{\left|{x - 1}\right|}{x}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x$$
$$\lim_{x \to -\infty}\left(\frac{\left|{x - 1}\right|}{x}\right) = -1$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = - x$$
$$\lim_{x \to \infty}\left(\frac{\left|{x - 1}\right|}{x}\right) = 1$$
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:
$$\left|{x - 1}\right| = \left|{x + 1}\right|$$
- No
$$\left|{x - 1}\right| = - \left|{x + 1}\right|$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\left|{x - 1}\right| = \left|{x + 1}\right|$$
- No
$$\left|{x - 1}\right| = - \left|{x + 1}\right|$$
- No
so, the function
not is
neither even, nor odd