What is sqrt(0.5x^2+1)<= 5|x| ?

The solution to sqrt(0.5x^2+1) = (sqrt(2))/7

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sqrt(0.5x^2+1)<= 5|x|

0.5 x^{2} + 1 \leq 5 ∣ x ∣

x \leq - \frac{2}{7} or x \geq \frac{2}{7}

Solution steps

0.5 x^{2} + 1 \leq 5 ∣ x ∣

f (x) \leq g (x)

then

g (x) \geq 0

5 ∣ x ∣ \geq 0 :

True for all

x

0.5 x^{2} + 1 \leq 5 ∣ x ∣ : x \leq - \frac{2}{7} or x \geq \frac{2}{7}

Verify Solutions:

x \leq - \frac{2}{7}

True

, x \geq \frac{2}{7}

True

The solution is

x \leq - \frac{2}{7} or x \geq \frac{2}{7}

\frac{- 3}{- 6}

r^{2} + 5 r + 4 = 0

∣ x ∣ = 3

- \frac{3}{8 ^{2}}

4 x^{2} = 6 - 5 x

What is sqrt(0.5x^2+1)<= 5|x| ?
The solution to sqrt(0.5x^2+1)<= 5|x| is x<=-(sqrt(2))/7 \lor x>= (sqrt(2))/7