Mister Exam
(2-x)*(4x+3)/(x-3)^3*(x+1)^2<=0 inequation
The solution
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(2 - x)*(4*x + 3) 2
-----------------*(x + 1) <= 0
3
(x - 3)
$$\frac{\left(2 - x\right) \left(4 x + 3\right)}{\left(x - 3\right)^{3}} \left(x + 1\right)^{2} \leq 0$$
(((2 - x)*(4*x + 3))/(x - 3)^3)*(x + 1)^2 <= 0
Detail solution
Given the inequality:
$$\frac{\left(2 - x\right) \left(4 x + 3\right)}{\left(x - 3\right)^{3}} \left(x + 1\right)^{2} \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\left(2 - x\right) \left(4 x + 3\right)}{\left(x - 3\right)^{3}} \left(x + 1\right)^{2} = 0$$
Solve:
$$x_{1} = -1$$
$$x_{2} = - \frac{3}{4}$$
$$x_{3} = 2$$
$$x_{1} = -1$$
$$x_{2} = - \frac{3}{4}$$
$$x_{3} = 2$$
This roots
$$x_{1} = -1$$
$$x_{2} = - \frac{3}{4}$$
$$x_{3} = 2$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-1 + - \frac{1}{10}$$
=
$$- \frac{11}{10}$$
substitute to the expression
$$\frac{\left(2 - x\right) \left(4 x + 3\right)}{\left(x - 3\right)^{3}} \left(x + 1\right)^{2} \leq 0$$
$$\frac{\left(2 - - \frac{11}{10}\right) \left(\frac{\left(-11\right) 4}{10} + 3\right)}{\left(-3 + - \frac{11}{10}\right)^{3}} \left(- \frac{11}{10} + 1\right)^{2} \leq 0$$
but
Then
$$x \leq -1$$
no execute
one of the solutions of our inequality is:
$$x \geq -1 \wedge x \leq - \frac{3}{4}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \geq -1 \wedge x \leq - \frac{3}{4}$$
$$x \geq 2$$
$$\frac{\left(2 - x\right) \left(4 x + 3\right)}{\left(x - 3\right)^{3}} \left(x + 1\right)^{2} \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\left(2 - x\right) \left(4 x + 3\right)}{\left(x - 3\right)^{3}} \left(x + 1\right)^{2} = 0$$
Solve:
$$x_{1} = -1$$
$$x_{2} = - \frac{3}{4}$$
$$x_{3} = 2$$
$$x_{1} = -1$$
$$x_{2} = - \frac{3}{4}$$
$$x_{3} = 2$$
This roots
$$x_{1} = -1$$
$$x_{2} = - \frac{3}{4}$$
$$x_{3} = 2$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-1 + - \frac{1}{10}$$
=
$$- \frac{11}{10}$$
substitute to the expression
$$\frac{\left(2 - x\right) \left(4 x + 3\right)}{\left(x - 3\right)^{3}} \left(x + 1\right)^{2} \leq 0$$
$$\frac{\left(2 - - \frac{11}{10}\right) \left(\frac{\left(-11\right) 4}{10} + 3\right)}{\left(-3 + - \frac{11}{10}\right)^{3}} \left(- \frac{11}{10} + 1\right)^{2} \leq 0$$
217 ------ <= 0 344605
but
217 ------ >= 0 344605
Then
$$x \leq -1$$
no execute
one of the solutions of our inequality is:
$$x \geq -1 \wedge x \leq - \frac{3}{4}$$
_____ _____
/ \ /
-------•-------•-------•-------
x1 x2 x3Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \geq -1 \wedge x \leq - \frac{3}{4}$$
$$x \geq 2$$
Solving inequality on a graph
Rapid solution
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Or(And(-3/4 <= x, x <= 2), And(3 < x, x < oo))
$$\left(- \frac{3}{4} \leq x \wedge x \leq 2\right) \vee \left(3 < x \wedge x < \infty\right)$$
((-3/4 <= x)∧(x <= 2))∨((3 < x)∧(x < oo))
Rapid solution 2
[src]
[-3/4, 2] U (3, oo)
$$x\ in\ \left[- \frac{3}{4}, 2\right] \cup \left(3, \infty\right)$$
x in Union(Interval(-3/4, 2), Interval.open(3, oo))