What is the range of sqrt(x^2-1/2) ?

The range of sqrt(x^2-1/2) is f(x)>= 0

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range of sqrt(x^2-1/2)

range

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

f (x) \geq 0

Interval Notation

[0, \infty)

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

x^{2} - \frac{1}{2} : x \leq - \frac{1}{2} or x \geq \frac{1}{2}

Extreme Points of

x^{2} - \frac{1}{2} :

Minimum

(- \frac{1}{2}, 0),

Minimum

(\frac{1}{2}, 0)

Find the range for the interval

- \infty < x \leq - \frac{1}{2} : 0 \leq f (x) < \infty

Find the range for the interval

\frac{1}{2} \leq x < \infty : 0 \leq f (x) < \infty

Combine the ranges of all domain intervals to obtain the function range

f (x) \geq 0 or f (x) \geq 0

f (x) \geq 0

x^{2} - 4

f (x) = x + 1 + x + 2

x^{3} + 4

f (x) = \frac{1}{6 2 x + 16 - 12}

f (x) = \frac{4}{x + 10}