Mister Exam
-2x^3+3x-1>0 inequation
The solution
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3 - 2*x + 3*x - 1 > 0
$$\left(- 2 x^{3} + 3 x\right) - 1 > 0$$
-2*x^3 + 3*x - 1 > 0
Detail solution
Given the inequality:
$$\left(- 2 x^{3} + 3 x\right) - 1 > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(- 2 x^{3} + 3 x\right) - 1 = 0$$
Solve:
$$x_{1} = 1$$
$$x_{2} = - \frac{\sqrt{3}}{2} - \frac{1}{2}$$
$$x_{3} = - \frac{1}{2} + \frac{\sqrt{3}}{2}$$
$$x_{1} = 1$$
$$x_{2} = - \frac{\sqrt{3}}{2} - \frac{1}{2}$$
$$x_{3} = - \frac{1}{2} + \frac{\sqrt{3}}{2}$$
This roots
$$x_{2} = - \frac{\sqrt{3}}{2} - \frac{1}{2}$$
$$x_{3} = - \frac{1}{2} + \frac{\sqrt{3}}{2}$$
$$x_{1} = 1$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$\left(- \frac{\sqrt{3}}{2} - \frac{1}{2}\right) + - \frac{1}{10}$$
=
$$- \frac{\sqrt{3}}{2} - \frac{3}{5}$$
substitute to the expression
$$\left(- 2 x^{3} + 3 x\right) - 1 > 0$$
$$-1 + \left(3 \left(- \frac{\sqrt{3}}{2} - \frac{3}{5}\right) - 2 \left(- \frac{\sqrt{3}}{2} - \frac{3}{5}\right)^{3}\right) > 0$$
one of the solutions of our inequality is:
$$x < - \frac{\sqrt{3}}{2} - \frac{1}{2}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - \frac{\sqrt{3}}{2} - \frac{1}{2}$$
$$x > - \frac{1}{2} + \frac{\sqrt{3}}{2} \wedge x < 1$$
$$\left(- 2 x^{3} + 3 x\right) - 1 > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(- 2 x^{3} + 3 x\right) - 1 = 0$$
Solve:
$$x_{1} = 1$$
$$x_{2} = - \frac{\sqrt{3}}{2} - \frac{1}{2}$$
$$x_{3} = - \frac{1}{2} + \frac{\sqrt{3}}{2}$$
$$x_{1} = 1$$
$$x_{2} = - \frac{\sqrt{3}}{2} - \frac{1}{2}$$
$$x_{3} = - \frac{1}{2} + \frac{\sqrt{3}}{2}$$
This roots
$$x_{2} = - \frac{\sqrt{3}}{2} - \frac{1}{2}$$
$$x_{3} = - \frac{1}{2} + \frac{\sqrt{3}}{2}$$
$$x_{1} = 1$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$\left(- \frac{\sqrt{3}}{2} - \frac{1}{2}\right) + - \frac{1}{10}$$
=
$$- \frac{\sqrt{3}}{2} - \frac{3}{5}$$
substitute to the expression
$$\left(- 2 x^{3} + 3 x\right) - 1 > 0$$
$$-1 + \left(3 \left(- \frac{\sqrt{3}}{2} - \frac{3}{5}\right) - 2 \left(- \frac{\sqrt{3}}{2} - \frac{3}{5}\right)^{3}\right) > 0$$
3
/ ___\ ___
14 | 3 \/ 3 | 3*\/ 3 > 0
- -- - 2*|- - - -----| - -------
5 \ 5 2 / 2 one of the solutions of our inequality is:
$$x < - \frac{\sqrt{3}}{2} - \frac{1}{2}$$
_____ _____
\ / \
-------ο-------ο-------ο-------
x2 x3 x1Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - \frac{\sqrt{3}}{2} - \frac{1}{2}$$
$$x > - \frac{1}{2} + \frac{\sqrt{3}}{2} \wedge x < 1$$
Solving inequality on a graph
Rapid solution 2
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___ ___
1 \/ 3 1 \/ 3
(-oo, - - - -----) U (- - + -----, 1)
2 2 2 2
$$x\ in\ \left(-\infty, - \frac{\sqrt{3}}{2} - \frac{1}{2}\right) \cup \left(- \frac{1}{2} + \frac{\sqrt{3}}{2}, 1\right)$$
x in Union(Interval.open(-oo, -sqrt(3)/2 - 1/2), Interval.open(-1/2 + sqrt(3)/2, 1))
Rapid solution
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/ / ___\ / ___ \\ | | 1 \/ 3 | | 1 \/ 3 || Or|And|-oo < x, x < - - - -----|, And|x < 1, - - + ----- < x|| \ \ 2 2 / \ 2 2 //
$$\left(-\infty < x \wedge x < - \frac{\sqrt{3}}{2} - \frac{1}{2}\right) \vee \left(x < 1 \wedge - \frac{1}{2} + \frac{\sqrt{3}}{2} < x\right)$$
((-oo < x)∧(x < -1/2 - sqrt(3)/2))∨((x < 1)∧(-1/2 + sqrt(3)/2 < x))