Solve ((x-3)(x-2))/([(x^3)(x^2)]) (-(x-1))/x ≥0

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

x \in (- \infty, 0) \cup (0, 1) \cup [2, 3]

Evaluate

((x - 3) (x - 2)) \div (x^{3} \times x^{2}) (- (x - 1)) \div x \geq 0

Find the domain

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((x - 3) (x - 2)) \div (x^{3} \times x^{2}) (- (x - 1)) \div x \geq 0, x \in (- \infty, 0) \cup (0, 1) \cup (1, + \infty)

Calculate

((x - 3) (x - 2)) \div (x^{3} \times x^{2} (- (x - 1))) \div x \geq 0

Simplify

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\frac{( x - 3 ) ( x - 2 )}{x ^{6} ( - x + 1 )} \geq 0

Set the numerator and denominator of \frac{( x - 3 ) ( x - 2 )}{x ^{6} ( - x + 1 )} equal to 0 to find the values of x where sign changes may occur

(x - 3) (x - 2) = 0 x^{6} (- x + 1) = 0

Calculate

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x = 3 x = 2 x^{6} (- x + 1) = 0

Calculate

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x = 3 x = 2 x = 0 x = 1

Determine the test intervals using the critical values

x < 0 0 < x < 1 1 < x < 2 2 < x < 3 x > 3

Choose a value form each interval

x_{1} = - 1 x_{2} = \frac{1}{2} x_{3} = \frac{3}{2} x_{4} = \frac{5}{2} x_{5} = 4

To determine if x < 0 is the solution to the inequality,test if the chosen value x = - 1 satisfies the initial inequality

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x < 0 is the solution x_{2} = \frac{1}{2} x_{3} = \frac{3}{2} x_{4} = \frac{5}{2} x_{5} = 4

To determine if 0 < x < 1 is the solution to the inequality,test if the chosen value x = \frac{1}{2} satisfies the initial inequality

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x < 0 is the solution 0 < x < 1 is the solution x_{3} = \frac{3}{2} x_{4} = \frac{5}{2} x_{5} = 4

To determine if 1 < x < 2 is the solution to the inequality,test if the chosen value x = \frac{3}{2} satisfies the initial inequality

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x < 0 is the solution 0 < x < 1 is the solution 1 < x < 2 is not a solution x_{4} = \frac{5}{2} x_{5} = 4

To determine if 2 < x < 3 is the solution to the inequality,test if the chosen value x = \frac{5}{2} satisfies the initial inequality

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x < 0 is the solution 0 < x < 1 is the solution 1 < x < 2 is not a solution 2 < x < 3 is the solution x_{5} = 4

To determine if x > 3 is the solution to the inequality,test if the chosen value x = 4 satisfies the initial inequality

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x < 0 is the solution 0 < x < 1 is the solution 1 < x < 2 is not a solution 2 < x < 3 is the solution x > 3 is not a solution

The original inequality is a nonstrict inequality,so include the critical value in the solution

x < 0 is the solution 0 < x < 1 is the solution 2 \leq x \leq 3 is the solution

The final solution of the original inequality is x \in (- \infty, 0) \cup (0, 1) \cup [2, 3]

x \in (- \infty, 0) \cup (0, 1) \cup [2, 3]

Check if the solution is in the defined range

x \in (- \infty, 0) \cup (0, 1) \cup [2, 3], x \in (- \infty, 0) \cup (0, 1) \cup (1, + \infty)

Solution

x \in (- \infty, 0) \cup (0, 1) \cup [2, 3]

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