Mister Exam
Graphing y = (x^4-3)/x
The solution
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4
x - 3
f(x) = ------
x
$$f{\left(x \right)} = \frac{x^{4} - 3}{x}$$
f = (x^4 - 3)/x
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 0$$
$$x_{1} = 0$$
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:
$$\frac{x^{4} - 3}{x} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = - \sqrt[4]{3}$$
$$x_{2} = \sqrt[4]{3}$$
Numerical solution
$$x_{1} = 1.31607401295249$$
$$x_{2} = -1.31607401295249$$
so we need to solve the equation:
$$\frac{x^{4} - 3}{x} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = - \sqrt[4]{3}$$
$$x_{2} = \sqrt[4]{3}$$
Numerical solution
$$x_{1} = 1.31607401295249$$
$$x_{2} = -1.31607401295249$$
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
$$4 x^{2} - \frac{x^{4} - 3}{x^{2}} = 0$$
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
$$4 x^{2} - \frac{x^{4} - 3}{x^{2}} = 0$$
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
$$2 \left(2 x + \frac{x^{4} - 3}{x^{3}}\right) = 0$$
Solve this equation
The roots of this equation
$$x_{1} = -1$$
$$x_{2} = 1$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(2 \left(2 x + \frac{x^{4} - 3}{x^{3}}\right)\right) = \infty$$
$$\lim_{x \to 0^+}\left(2 \left(2 x + \frac{x^{4} - 3}{x^{3}}\right)\right) = -\infty$$
- the limits are not equal, so
$$x_{1} = 0$$
- is an inflection point
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
$$\left[1, \infty\right)$$
Convex at the intervals
$$\left(-\infty, -1\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$2 \left(2 x + \frac{x^{4} - 3}{x^{3}}\right) = 0$$
Solve this equation
The roots of this equation
$$x_{1} = -1$$
$$x_{2} = 1$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(2 \left(2 x + \frac{x^{4} - 3}{x^{3}}\right)\right) = \infty$$
$$\lim_{x \to 0^+}\left(2 \left(2 x + \frac{x^{4} - 3}{x^{3}}\right)\right) = -\infty$$
- the limits are not equal, so
$$x_{1} = 0$$
- is an inflection point
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
$$\left[1, \infty\right)$$
Convex at the intervals
$$\left(-\infty, -1\right]$$
Vertical asymptotes
Have:
$$x_{1} = 0$$
$$x_{1} = 0$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{x^{4} - 3}{x}\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{x^{4} - 3}{x}\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\frac{x^{4} - 3}{x}\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{x^{4} - 3}{x}\right) = \infty$$
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^4 - 3)/x, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{x^{4} - 3}{x^{2}}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{x^{4} - 3}{x^{2}}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\frac{x^{4} - 3}{x^{2}}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{x^{4} - 3}{x^{2}}\right) = \infty$$
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:
$$\frac{x^{4} - 3}{x} = - \frac{x^{4} - 3}{x}$$
- No
$$\frac{x^{4} - 3}{x} = \frac{x^{4} - 3}{x}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{x^{4} - 3}{x} = - \frac{x^{4} - 3}{x}$$
- No
$$\frac{x^{4} - 3}{x} = \frac{x^{4} - 3}{x}$$
- No
so, the function
not is
neither even, nor odd