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  • Graphing y =:
  • 3x^4-4x^3
  • 2x^2+5x+2
  • 2x^2-5x
  • (2x^2-1)/x
  • Identical expressions

  • (two x^2- one)/x
  • (2x squared minus 1) divide by x
  • (two x squared minus one) divide by x
  • (2x2-1)/x
  • 2x2-1/x
  • (2x²-1)/x
  • (2x to the power of 2-1)/x
  • 2x^2-1/x
  • (2x^2-1) divide by x
  • Similar expressions

  • (2x^2+1)/x

Graphing y = (2x^2-1)/x

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src]
          2    
       2*x  - 1
f(x) = --------
          x    
$$f{\left(x \right)} = \frac{2 x^{2} - 1}{x}$$
f = (2*x^2 - 1)/x
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 0$$
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:
$$\frac{2 x^{2} - 1}{x} = 0$$
Solve this equation
The points of intersection with the axis X:

Analytical solution
$$x_{1} = - \frac{\sqrt{2}}{2}$$
$$x_{2} = \frac{\sqrt{2}}{2}$$
Numerical solution
$$x_{1} = -0.707106781186548$$
$$x_{2} = 0.707106781186548$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (2*x^2 - 1)/x.
$$\frac{-1 + 2 \cdot 0^{2}}{0}$$
The result:
$$f{\left(0 \right)} = \tilde{\infty}$$
sof doesn't intersect Y
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
$$4 - \frac{2 x^{2} - 1}{x^{2}} = 0$$
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
$$\frac{2 \left(-2 + \frac{2 x^{2} - 1}{x^{2}}\right)}{x} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
$$x_{1} = 0$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{2 x^{2} - 1}{x}\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{2 x^{2} - 1}{x}\right) = \infty$$
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 (2*x^2 - 1)/x, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{2 x^{2} - 1}{x^{2}}\right) = 2$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = 2 x$$
$$\lim_{x \to \infty}\left(\frac{2 x^{2} - 1}{x^{2}}\right) = 2$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = 2 x$$
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:
$$\frac{2 x^{2} - 1}{x} = - \frac{2 x^{2} - 1}{x}$$
- No
$$\frac{2 x^{2} - 1}{x} = \frac{2 x^{2} - 1}{x}$$
- No
so, the function
not is
neither even, nor odd