What is sqrt(1-x)<= \sqrt[4]{x+5} ?

The solution to sqrt(1-x)<= \sqrt[4]{x+5} is -1<= x<= 1

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sqrt(1-x)<= \sqrt[4]{x+5}

- 1 \leq x \leq 1

Interval Notation

[- 1, 1]

Solution steps

f (x) \leq g (x)

then

g (x) \geq 0

True for all

x

Find singularity points

Find non-negative values for radicals:

- 5 \leq x \leq 1

Combine the intervals

True for all x and x \geq - 1 and - 5 \leq x \leq 1

Merge Overlapping Intervals

- 1 \leq x \leq 1

\frac{y ^{\frac{5}{4}}}{y ^{\frac{1}{4}}}

22 x y - 15 x^{2} - 8 y^{2}

4 ∣ z - 5 ∣ - 9 > 14

x^{2} + 11 x + 30 = 0

x^{3} + 3 x^{2} + 5 x

What is sqrt(1-x)<= \sqrt[4]{x+5} ?
The solution to sqrt(1-x)<= \sqrt[4]{x+5} is -1<= x<= 1