What is x+sqrt(x^2+1)>0 ?

The solution to x+sqrt(x^2+1)>0 is True for all x

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x+sqrt(x^2+1)>0

x + x^{2} + 1 > 0

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

x + x^{2} + 1 > 0

Subtract

x

from both sides

x + x^{2} + 1 - x > 0 - x

Simplify

x^{2} + 1 > - x

f (x) > g (x)

then

g (x) < 0 or g (x) \geq 0

For

- x < 0 : x > 0

For

- x \geq 0 : x \leq 0

Combine the intervals

x \leq 0 or x > 0

Merge Overlapping Intervals

True for all x

Verify Solutions:

The solution is

True for all x

x^{2} + 3 x - 10 > 0

- x^{2} + 14 x - 49 = 0

lo g_{\frac{1}{4}} (64) = n

49 - 4 z^{2}

(4 + \frac{1}{2})^{2}