What is 3-x>3(1-x^2)^{1/2} ?

The solution to 3-x>3(1-x^2)^{1/2} is -1<= x<0\lor 3/5 <x<= 1

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3-x>3(1-x^2)^{1/2}

3 - x > 3 (1 - x^{2})^{\frac{1}{2}}

- 1 \leq x < 0 or \frac{3}{5} < x \leq 1

Interval Notation

[- 1, 0) \cup (\frac{3}{5}, 1]

Decimal Notation

- 1 \leq x < 0 or 0.6 < x \leq 1

Solution steps

3 - x > 3 (1 - x^{2})^{\frac{1}{2}}

Switch sides

3 (1 - x^{2})^{\frac{1}{2}} < 3 - x

Simplify

3 (1 - x^{2})^{\frac{1}{2}} < 3 - x : 3 1 - x^{2} < 3 - x

3 1 - x^{2} < 3 - x

Divide both sides by

3

\frac{3 1 - x ^{2}}{3} < \frac{3}{3} - \frac{x}{3}

Simplify

1 - x^{2} < \frac{3 - x}{3}

f (x) < g (x)

then

g (x) > 0

\frac{3 - x}{3} > 0 : x < 3

1 - x^{2} < \frac{3 - x}{3} : x < 0 or x > \frac{3}{5}

Find singularity points

Find non-negative values for radicals:

- 1 \leq x \leq 1

Combine the intervals

x < 3 and (x < 0 or x > \frac{3}{5}) and - 1 \leq x \leq 1

Merge Overlapping Intervals

- 1 \leq x < 0 or \frac{3}{5} < x \leq 1

Verify Solutions:

- 1 \leq x < 0

True

, \frac{3}{5} < x \leq 1

True

The solution is

- 1 \leq x < 0 or \frac{3}{5} < x \leq 1

x + 5 - 25 - 10 x + x^{2} \leq 2 x

2 x \leq x^{2}

x - 4 < - 3

x^{2} - 4 x + 3 \geq 2 - x

(x - 1) \cdot x^{2} - x - 2 \geq 0

What is 3-x>3(1-x^2)^{1/2} ?
The solution to 3-x>3(1-x^2)^{1/2} is -1<= x<0\lor 3/5 <x<= 1