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

Graphing y = x^3+3x-1

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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        3          
f(x) = x  + 3*x - 1
f(x)=(x3+3x)−1f{\left(x \right)} = \left(x^{3} + 3 x\right) - 1 
f = x^3 + 3*x - 1
The graph of the function
02468-8-6-4-2-1010-20002000
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:
(x3+3x)−1=0\left(x^{3} + 3 x\right) - 1 = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=−112+523+12+523x_{1} = - \frac{1}{\sqrt[3]{\frac{1}{2} + \frac{\sqrt{5}}{2}}} + \sqrt[3]{\frac{1}{2} + \frac{\sqrt{5}}{2}}
Numerical solution
x1=0.322185354626086x_{1} = 0.322185354626086
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*x - 1.
−1+(03+0⋅3)-1 + \left(0^{3} + 0 \cdot 3\right)
The result:
f(0)=−1f{\left(0 \right)} = -1
The point:
(0, -1)
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
3x2+3=03 x^{2} + 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
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
6x=06 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[0, \infty\right)
Convex at the intervals
(−∞,0]\left(-\infty, 0\right]
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞((x3+3x)−1)=−∞\lim_{x \to -\infty}\left(\left(x^{3} + 3 x\right) - 1\right) = -\infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞((x3+3x)−1)=∞\lim_{x \to \infty}\left(\left(x^{3} + 3 x\right) - 1\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*x - 1, divided by x at x->+oo and x ->-oo
lim⁡x→−∞((x3+3x)−1x)=∞\lim_{x \to -\infty}\left(\frac{\left(x^{3} + 3 x\right) - 1}{x}\right) = \infty
Let's take the limit
so,
inclined asymptote on the left doesn’t exist
lim⁡x→∞((x3+3x)−1x)=∞\lim_{x \to \infty}\left(\frac{\left(x^{3} + 3 x\right) - 1}{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:
(x3+3x)−1=−x3−3x−1\left(x^{3} + 3 x\right) - 1 = - x^{3} - 3 x - 1
- No
(x3+3x)−1=x3+3x+1\left(x^{3} + 3 x\right) - 1 = x^{3} + 3 x + 1
- No
so, the function
not is
neither even, nor odd
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