Mister Exam
Graphing y = 1/3x^3-2x+3x-4
The solution
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3
x
f(x) = -- - 2*x + 3*x - 4
3
$$f{\left(x \right)} = \left(3 x + \left(\frac{x^{3}}{3} - 2 x\right)\right) - 4$$
f = 3*x + x^3/3 - 2*x - 4
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:
$$\left(3 x + \left(\frac{x^{3}}{3} - 2 x\right)\right) - 4 = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = - \frac{1}{\sqrt[3]{6 + \sqrt{37}}} + \sqrt[3]{6 + \sqrt{37}}$$
Numerical solution
$$x_{1} = 1.85888907187124$$
so we need to solve the equation:
$$\left(3 x + \left(\frac{x^{3}}{3} - 2 x\right)\right) - 4 = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = - \frac{1}{\sqrt[3]{6 + \sqrt{37}}} + \sqrt[3]{6 + \sqrt{37}}$$
Numerical solution
$$x_{1} = 1.85888907187124$$
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
$$x^{2} + 1 = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(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
$$x^{2} + 1 = 0$$
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
$$2 x = 0$$
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
$$\left(-\infty, 0\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$2 x = 0$$
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
$$\lim_{x \to -\infty}\left(\left(3 x + \left(\frac{x^{3}}{3} - 2 x\right)\right) - 4\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\left(3 x + \left(\frac{x^{3}}{3} - 2 x\right)\right) - 4\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\left(3 x + \left(\frac{x^{3}}{3} - 2 x\right)\right) - 4\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\left(3 x + \left(\frac{x^{3}}{3} - 2 x\right)\right) - 4\right) = \infty$$
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 - 2*x + 3*x - 4, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(3 x + \left(\frac{x^{3}}{3} - 2 x\right)\right) - 4}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{\left(3 x + \left(\frac{x^{3}}{3} - 2 x\right)\right) - 4}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\frac{\left(3 x + \left(\frac{x^{3}}{3} - 2 x\right)\right) - 4}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{\left(3 x + \left(\frac{x^{3}}{3} - 2 x\right)\right) - 4}{x}\right) = \infty$$
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:
$$\left(3 x + \left(\frac{x^{3}}{3} - 2 x\right)\right) - 4 = - \frac{x^{3}}{3} - x - 4$$
- No
$$\left(3 x + \left(\frac{x^{3}}{3} - 2 x\right)\right) - 4 = \frac{x^{3}}{3} + x + 4$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\left(3 x + \left(\frac{x^{3}}{3} - 2 x\right)\right) - 4 = - \frac{x^{3}}{3} - x - 4$$
- No
$$\left(3 x + \left(\frac{x^{3}}{3} - 2 x\right)\right) - 4 = \frac{x^{3}}{3} + x + 4$$
- No
so, the function
not is
neither even, nor odd