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  • Graphing y =:
  • 3x^2-12x+1
  • 2x^2+5x+2
  • (2x^2-1)/x
  • (1/x)+4x^2

  • 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
Plotting a function graph/ (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(x)=2x2−1xf{\left(x \right)} = \frac{2 x^{2} - 1}{x} 
f = (2*x^2 - 1)/x
The graph of the function
02468-8-6-4-2-1010-5050
The domain of the function
The points at which the function is not precisely defined:
x1=0x_{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:
2x2−1x=0\frac{2 x^{2} - 1}{x} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=−22x_{1} = - \frac{\sqrt{2}}{2}
x2=22x_{2} = \frac{\sqrt{2}}{2}
Numerical solution
x1=−0.707106781186548x_{1} = -0.707106781186548
x2=0.707106781186548x_{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.
−1+2⋅020\frac{-1 + 2 \cdot 0^{2}}{0}
The result:
f(0)=∞~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
ddxf(x)=0\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:
ddxf(x)=\frac{d}{d x} f{\left(x \right)} =
the first derivative
4−2x2−1x2=04 - \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
d2dx2f(x)=0\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:
d2dx2f(x)=\frac{d^{2}}{d x^{2}} f{\left(x \right)} =
the second derivative
2(−2+2x2−1x2)x=0\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:
x1=0x_{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→−∞(2x2−1x)=−∞\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→∞(2x2−1x)=∞\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→−∞(2x2−1x2)=2\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=2xy = 2 x
lim⁡x→∞(2x2−1x2)=2\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=2xy = 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:
2x2−1x=−2x2−1x\frac{2 x^{2} - 1}{x} = - \frac{2 x^{2} - 1}{x}
- No
2x2−1x=2x2−1x\frac{2 x^{2} - 1}{x} = \frac{2 x^{2} - 1}{x}
- No
so, the function
not is
neither even, nor odd
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