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-1/3x^3-4x
Plotting a function graph/ -1/3x^3-4x

Graphing y = -1/3x^3-4x

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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          3      
         x       
f(x) = - -- - 4*x
         3
f(x)=−x33−4xf{\left(x \right)} = - \frac{x^{3}}{3} - 4 x 
f = -x^3/3 - 4*x
The graph of the function
02468-8-6-4-2-1010-10001000
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:
−x33−4x=0- \frac{x^{3}}{3} - 4 x = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=0x_{1} = 0
Numerical solution
x1=0x_{1} = 0
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to -x^3/3 - 4*x.
−033−4⋅0- \frac{0^{3}}{3} - 4 \cdot 0
The result:
f(0)=0f{\left(0 \right)} = 0
The point:
(0, 0)
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
−x2−4=0- x^{2} - 4 = 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
−2x=0- 2 x = 0
Solve this equation
The roots of this equation
x1=0x_{1} = 0

Сonvexity and concavity intervals:
Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Concave at the intervals
(−∞,0]\left(-\infty, 0\right]
Convex at the intervals
[0,∞)\left[0, \infty\right)
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(−x33−4x)=∞\lim_{x \to -\infty}\left(- \frac{x^{3}}{3} - 4 x\right) = \infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞(−x33−4x)=−∞\lim_{x \to \infty}\left(- \frac{x^{3}}{3} - 4 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 -x^3/3 - 4*x, divided by x at x->+oo and x ->-oo
lim⁡x→−∞(−x33−4xx)=−∞\lim_{x \to -\infty}\left(\frac{- \frac{x^{3}}{3} - 4 x}{x}\right) = -\infty
Let's take the limit
so,
inclined asymptote on the left doesn’t exist
lim⁡x→∞(−x33−4xx)=−∞\lim_{x \to \infty}\left(\frac{- \frac{x^{3}}{3} - 4 x}{x}\right) = -\infty
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
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:
−x33−4x=x33+4x- \frac{x^{3}}{3} - 4 x = \frac{x^{3}}{3} + 4 x
- No
−x33−4x=−x33−4x- \frac{x^{3}}{3} - 4 x = - \frac{x^{3}}{3} - 4 x
- No
so, the function
not is
neither even, nor odd
The graph
Graphing y = -1/3x^3-4x
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