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1/(y^3)
Plotting a function graph/ 1/(y^3)

Graphing y = 1/(y^3)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
         1 
f(y) = 1*--
          3
         y
f(y)=1⋅1y3f{\left(y \right)} = 1 \cdot \frac{1}{y^{3}} 
f = 1/y^3
The graph of the function
02468-8-6-4-2-1010-2000020000
The domain of the function
The points at which the function is not precisely defined:
y1=0y_{1} = 0
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis Y at f = 0
so we need to solve the equation:
1⋅1y3=01 \cdot \frac{1}{y^{3}} = 0
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis Y
The points of intersection with the Y axis coordinate
The graph crosses Y axis when y equals 0:
substitute y = 0 to 1/y^3.
1⋅1031 \cdot \frac{1}{0^{3}}
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
ddyf(y)=0\frac{d}{d y} f{\left(y \right)} = 0
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
ddyf(y)=\frac{d}{d y} f{\left(y \right)} =
the first derivative
−3yy3=0- \frac{3}{y y^{3}} = 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
d2dy2f(y)=0\frac{d^{2}}{d y^{2}} f{\left(y \right)} = 0
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
d2dy2f(y)=\frac{d^{2}}{d y^{2}} f{\left(y \right)} =
the second derivative
12y5=0\frac{12}{y^{5}} = 0
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
y1=0y_{1} = 0
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at y->+oo and y->-oo
lim⁡y→−∞(1⋅1y3)=0\lim_{y \to -\infty}\left(1 \cdot \frac{1}{y^{3}}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡y→∞(1⋅1y3)=0\lim_{y \to \infty}\left(1 \cdot \frac{1}{y^{3}}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=0y = 0
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 1/y^3, divided by y at y->+oo and y ->-oo
lim⁡y→−∞(1yy3)=0\lim_{y \to -\infty}\left(\frac{1}{y y^{3}}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡y→∞(1yy3)=0\lim_{y \to \infty}\left(\frac{1}{y y^{3}}\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(-y) и f = -f(-y).
So, check:
1⋅1y3=−1y31 \cdot \frac{1}{y^{3}} = - \frac{1}{y^{3}}
- No
1⋅1y3=1y31 \cdot \frac{1}{y^{3}} = \frac{1}{y^{3}}
- No
so, the function
not is
neither even, nor odd
The graph
Graphing y = 1/(y^3)
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