Mister Exam

Graphing y = x^3+4x

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = x  + 4*x
$$f{\left(x \right)} = x^{3} + 4 x$$
f = x^3 + 4*x
The graph of the function
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:
$$x^{3} + 4 x = 0$$
Solve this equation
The points of intersection with the axis X:

Analytical solution
$$x_{1} = 0$$
Numerical solution
$$x_{1} = 0$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to x^3 + 4*x.
$$0^{3} + 0 \cdot 4$$
The result:
$$f{\left(0 \right)} = 0$$
The point:
(0, 0)
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
$$3 x^{2} + 4 = 0$$
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
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$6 x = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 0$$

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
$$\left[0, \infty\right)$$
Convex at the intervals
$$\left(-\infty, 0\right]$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(x^{3} + 4 x\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(x^{3} + 4 x\right) = \infty$$
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 + 4*x, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{x^{3} + 4 x}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{x^{3} + 4 x}{x}\right) = \infty$$
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:
$$x^{3} + 4 x = - x^{3} - 4 x$$
- No
$$x^{3} + 4 x = x^{3} + 4 x$$
- Yes
so, the function
is
odd