What is 2x-x^2>= 1x ?

The solution to 2x-x^2>= 1x is 0<= x<= 1

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2x-x^2>= 1x

2 x - x^{2} \geq 1 x

0 \leq x \leq 1

Interval Notation

[0, 1]

Solution steps

2 x - x^{2} \geq 1 \cdot x

Rewrite in standard form

- x^{2} + x \geq 0

Factor

- x^{2} + x : - x (x - 1)

- x (x - 1) \geq 0

Multiply both sides by

- 1

(reverse the inequality)

(- x (x - 1)) (- 1) \leq 0 \cdot (- 1)

Simplify

x (x - 1) \leq 0

Identify the intervals

0 \leq x \leq 1

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

x^{2} + 20 x + 100 < 0

\frac{( x ^{2} + x - 2 )}{( x - 4 )} \leq 0

(x - 7) (x - 8) < 0

x^{2} - 4 x + 6 \leq 0