Mister Exam
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-
How to use it?
- Graphing y =:
- -x^4+8x^2-16
- x^4-6x^3+9x^2
-
x^3+3x^2-24x+28
- x^2(x-7)
- Integral of d{x}:
-
x^3+x^(1/3)
-
Identical expressions
- x^ three +x^(one / three)
- x cubed plus x to the power of (1 divide by 3)
- x to the power of three plus x to the power of (one divide by three)
- x3+x(1/3)
- x3+x1/3
- x³+x^(1/3)
- x to the power of 3+x to the power of (1/3)
- x^3+x^1/3
- x^3+x^(1 divide by 3)
-
Similar expressions
- x^3-x^(1/3)
Graphing y = x^3+x^(1/3)
The solution
You have entered
[src]
3 3 ___ f(x) = x + \/ x
$$f{\left(x \right)} = \sqrt[3]{x} + x^{3}$$
f = x^(1/3) + x^3
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:
$$\sqrt[3]{x} + x^{3} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 0$$
Numerical solution
$$x_{1} = 0$$
so we need to solve the equation:
$$\sqrt[3]{x} + x^{3} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 0$$
Numerical solution
$$x_{1} = 0$$
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
$$3 x^{2} + \frac{1}{3 x^{\frac{2}{3}}} = 0$$
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
$$3 x^{2} + \frac{1}{3 x^{\frac{2}{3}}} = 0$$
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
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$2 \left(3 x - \frac{1}{9 x^{\frac{5}{3}}}\right) = 0$$
Solve this equation
The roots of this equation
$$x_{1} = \frac{3^{\frac{7}{8}}}{9}$$
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
$$\left[\frac{3^{\frac{7}{8}}}{9}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, \frac{3^{\frac{7}{8}}}{9}\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$2 \left(3 x - \frac{1}{9 x^{\frac{5}{3}}}\right) = 0$$
Solve this equation
The roots of this equation
$$x_{1} = \frac{3^{\frac{7}{8}}}{9}$$
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
$$\left[\frac{3^{\frac{7}{8}}}{9}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, \frac{3^{\frac{7}{8}}}{9}\right]$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\sqrt[3]{x} + x^{3}\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\sqrt[3]{x} + x^{3}\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\sqrt[3]{x} + x^{3}\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\sqrt[3]{x} + x^{3}\right) = \infty$$
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 + x^(1/3), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\sqrt[3]{x} + x^{3}}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{\sqrt[3]{x} + x^{3}}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\frac{\sqrt[3]{x} + x^{3}}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{\sqrt[3]{x} + x^{3}}{x}\right) = \infty$$
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:
$$\sqrt[3]{x} + x^{3} = - x^{3} + \sqrt[3]{- x}$$
- No
$$\sqrt[3]{x} + x^{3} = x^{3} - \sqrt[3]{- x}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\sqrt[3]{x} + x^{3} = - x^{3} + \sqrt[3]{- x}$$
- No
$$\sqrt[3]{x} + x^{3} = x^{3} - \sqrt[3]{- x}$$
- No
so, the function
not is
neither even, nor odd