What is sqrt(0.5x^2+1)<= 4|x| ?

The solution to sqrt(0.5x^2+1) = sqrt(2/31)

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sqrt(0.5x^2+1)<= 4|x|

0.5 x^{2} + 1 \leq 4 ∣ x ∣

x \leq - \frac{2}{31} or x \geq \frac{2}{31}

Solution steps

0.5 x^{2} + 1 \leq 4 ∣ x ∣

f (x) \leq g (x)

then

g (x) \geq 0

4 ∣ x ∣ \geq 0 :

True for all

x

0.5 x^{2} + 1 \leq 4 ∣ x ∣ : x \leq - \frac{2}{31} or x \geq \frac{2}{31}

Verify Solutions:

x \leq - \frac{2}{31}

True

, x \geq \frac{2}{31}

True

The solution is

x \leq - \frac{2}{31} or x \geq \frac{2}{31}

\frac{1}{4} (1 - 3 x) = - 2

m^{2} - n^{2}

3 x (x + 6) = 2 (3 x - 10)

∣ 2 s - 7 ∣ > - 1

3 (x + 2) = 15

What is sqrt(0.5x^2+1)<= 4|x| ?
The solution to sqrt(0.5x^2+1)<= 4|x| is x<=-sqrt(2/31)\lor x>= sqrt(2/31)