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How to use it?
- Graphing y =:
- x(2-x)^2
-
|x|x-|x|-3x
- (x-3)/(x-1)
- x^3+x^2
-
Identical expressions
- ((x- two)*(x- four)*(x- eight)*(x- two hundred and fifty-six))/((x- one)*(x- three)*(x- seven)*(x- two hundred and fifty-five))
- ((x minus 2) multiply by (x minus 4) multiply by (x minus 8) multiply by (x minus 256)) divide by ((x minus 1) multiply by (x minus 3) multiply by (x minus 7) multiply by (x minus 255))
- ((x minus two) multiply by (x minus four) multiply by (x minus eight) multiply by (x minus two hundred and fifty minus six)) divide by ((x minus one) multiply by (x minus three) multiply by (x minus seven) multiply by (x minus two hundred and fifty minus five))
- ((x-2)(x-4)(x-8)(x-256))/((x-1)(x-3)(x-7)(x-255))
- x-2x-4x-8x-256/x-1x-3x-7x-255
- ((x-2)*(x-4)*(x-8)*(x-256)) divide by ((x-1)*(x-3)*(x-7)*(x-255))
-
Similar expressions
- ((x-2)*(x-4)*(x-8)*(x-256))/((x-1)*(x-3)*(x+7)*(x-255))
- ((x-2)*(x-4)*(x-8)*(x+256))/((x-1)*(x-3)*(x-7)*(x-255))
- ((x-2)*(x-4)*(x-8)*(x-256))/((x-1)*(x-3)*(x-7)*(x+255))
- ((x-2)*(x-4)*(x+8)*(x-256))/((x-1)*(x-3)*(x-7)*(x-255))
- ((x-2)*(x+4)*(x-8)*(x-256))/((x-1)*(x-3)*(x-7)*(x-255))
- ((x-2)*(x-4)*(x-8)*(x-256))/((x-1)*(x+3)*(x-7)*(x-255))
- ((x+2)*(x-4)*(x-8)*(x-256))/((x-1)*(x-3)*(x-7)*(x-255))
- ((x-2)*(x-4)*(x-8)*(x-256))/((x+1)*(x-3)*(x-7)*(x-255))
Graphing y = ((x-2)*(x-4)*(x-8)*(x-256))/((x-1)*(x-3)*(x-7)*(x-255))
The solution
You have entered
[src]
(x - 2)*(x - 4)*(x - 8)*(x - 256)
f(x) = ---------------------------------
(x - 1)*(x - 3)*(x - 7)*(x - 255)
$$f{\left(x \right)} = \frac{\left(x - 4\right) \left(x - 2\right) \left(x - 8\right) \left(x - 256\right)}{\left(x - 3\right) \left(x - 1\right) \left(x - 7\right) \left(x - 255\right)}$$
f = ((((x - 4)*(x - 2))*(x - 8))*(x - 256))/(((((x - 3)*(x - 1))*(x - 7))*(x - 255)))
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 1$$
$$x_{2} = 3$$
$$x_{3} = 7$$
$$x_{4} = 255$$
$$x_{1} = 1$$
$$x_{2} = 3$$
$$x_{3} = 7$$
$$x_{4} = 255$$
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:
$$\frac{\left(x - 4\right) \left(x - 2\right) \left(x - 8\right) \left(x - 256\right)}{\left(x - 3\right) \left(x - 1\right) \left(x - 7\right) \left(x - 255\right)} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 2$$
$$x_{2} = 4$$
$$x_{3} = 8$$
$$x_{4} = 256$$
Numerical solution
$$x_{1} = 2$$
$$x_{2} = 4$$
$$x_{3} = 8$$
$$x_{4} = 256$$
so we need to solve the equation:
$$\frac{\left(x - 4\right) \left(x - 2\right) \left(x - 8\right) \left(x - 256\right)}{\left(x - 3\right) \left(x - 1\right) \left(x - 7\right) \left(x - 255\right)} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 2$$
$$x_{2} = 4$$
$$x_{3} = 8$$
$$x_{4} = 256$$
Numerical solution
$$x_{1} = 2$$
$$x_{2} = 4$$
$$x_{3} = 8$$
$$x_{4} = 256$$
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
$$\frac{\left(x - 256\right) \left(x - 8\right) \left(x - 4\right) \left(x - 2\right) \left(- \left(x - 3\right) \left(x - 1\right) \left(x - 7\right) - \left(x - 255\right) \left(\left(x - 7\right) \left(2 x - 4\right) + \left(x - 3\right) \left(x - 1\right)\right)\right)}{\left(x - 255\right)^{2} \left(x - 7\right)^{2} \left(x - 3\right)^{2} \left(x - 1\right)^{2}} + \frac{1}{\left(x - 255\right) \left(x - 7\right) \left(x - 3\right) \left(x - 1\right)} \left(\left(x - 4\right) \left(x - 2\right) \left(x - 8\right) + \left(x - 256\right) \left(\left(x - 8\right) \left(2 x - 6\right) + \left(x - 4\right) \left(x - 2\right)\right)\right) = 0$$
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
$$\frac{\left(x - 256\right) \left(x - 8\right) \left(x - 4\right) \left(x - 2\right) \left(- \left(x - 3\right) \left(x - 1\right) \left(x - 7\right) - \left(x - 255\right) \left(\left(x - 7\right) \left(2 x - 4\right) + \left(x - 3\right) \left(x - 1\right)\right)\right)}{\left(x - 255\right)^{2} \left(x - 7\right)^{2} \left(x - 3\right)^{2} \left(x - 1\right)^{2}} + \frac{1}{\left(x - 255\right) \left(x - 7\right) \left(x - 3\right) \left(x - 1\right)} \left(\left(x - 4\right) \left(x - 2\right) \left(x - 8\right) + \left(x - 256\right) \left(\left(x - 8\right) \left(2 x - 6\right) + \left(x - 4\right) \left(x - 2\right)\right)\right) = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
Vertical asymptotes
Have:
$$x_{1} = 1$$
$$x_{2} = 3$$
$$x_{3} = 7$$
$$x_{4} = 255$$
$$x_{1} = 1$$
$$x_{2} = 3$$
$$x_{3} = 7$$
$$x_{4} = 255$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(x - 4\right) \left(x - 2\right) \left(x - 8\right) \left(x - 256\right)}{\left(x - 3\right) \left(x - 1\right) \left(x - 7\right) \left(x - 255\right)}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{\left(x - 4\right) \left(x - 2\right) \left(x - 8\right) \left(x - 256\right)}{\left(x - 3\right) \left(x - 1\right) \left(x - 7\right) \left(x - 255\right)}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
$$\lim_{x \to -\infty}\left(\frac{\left(x - 4\right) \left(x - 2\right) \left(x - 8\right) \left(x - 256\right)}{\left(x - 3\right) \left(x - 1\right) \left(x - 7\right) \left(x - 255\right)}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{\left(x - 4\right) \left(x - 2\right) \left(x - 8\right) \left(x - 256\right)}{\left(x - 3\right) \left(x - 1\right) \left(x - 7\right) \left(x - 255\right)}\right) = 1$$
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 ((((x - 2)*(x - 4))*(x - 8))*(x - 256))/(((((x - 1)*(x - 3))*(x - 7))*(x - 255))), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\frac{1}{\left(x - 255\right) \left(x - 7\right) \left(x - 3\right) \left(x - 1\right)} \left(x - 256\right) \left(x - 8\right) \left(x - 4\right) \left(x - 2\right)}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\frac{1}{\left(x - 255\right) \left(x - 7\right) \left(x - 3\right) \left(x - 1\right)} \left(x - 256\right) \left(x - 8\right) \left(x - 4\right) \left(x - 2\right)}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\frac{1}{\left(x - 255\right) \left(x - 7\right) \left(x - 3\right) \left(x - 1\right)} \left(x - 256\right) \left(x - 8\right) \left(x - 4\right) \left(x - 2\right)}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\frac{1}{\left(x - 255\right) \left(x - 7\right) \left(x - 3\right) \left(x - 1\right)} \left(x - 256\right) \left(x - 8\right) \left(x - 4\right) \left(x - 2\right)}{x}\right) = 0$$
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:
$$\frac{\left(x - 4\right) \left(x - 2\right) \left(x - 8\right) \left(x - 256\right)}{\left(x - 3\right) \left(x - 1\right) \left(x - 7\right) \left(x - 255\right)} = \frac{\left(- x - 256\right) \left(- x - 8\right) \left(- x - 4\right) \left(- x - 2\right)}{\left(- x - 255\right) \left(- x - 7\right) \left(- x - 3\right) \left(- x - 1\right)}$$
- No
$$\frac{\left(x - 4\right) \left(x - 2\right) \left(x - 8\right) \left(x - 256\right)}{\left(x - 3\right) \left(x - 1\right) \left(x - 7\right) \left(x - 255\right)} = - \frac{\left(- x - 256\right) \left(- x - 8\right) \left(- x - 4\right) \left(- x - 2\right)}{\left(- x - 255\right) \left(- x - 7\right) \left(- x - 3\right) \left(- x - 1\right)}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{\left(x - 4\right) \left(x - 2\right) \left(x - 8\right) \left(x - 256\right)}{\left(x - 3\right) \left(x - 1\right) \left(x - 7\right) \left(x - 255\right)} = \frac{\left(- x - 256\right) \left(- x - 8\right) \left(- x - 4\right) \left(- x - 2\right)}{\left(- x - 255\right) \left(- x - 7\right) \left(- x - 3\right) \left(- x - 1\right)}$$
- No
$$\frac{\left(x - 4\right) \left(x - 2\right) \left(x - 8\right) \left(x - 256\right)}{\left(x - 3\right) \left(x - 1\right) \left(x - 7\right) \left(x - 255\right)} = - \frac{\left(- x - 256\right) \left(- x - 8\right) \left(- x - 4\right) \left(- x - 2\right)}{\left(- x - 255\right) \left(- x - 7\right) \left(- x - 3\right) \left(- x - 1\right)}$$
- No
so, the function
not is
neither even, nor odd