What is the range of sqrt(x^2+4) ?

The range of sqrt(x^2+4) is f(x)>= 2

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range of sqrt(x^2+4)

range

x^{2} + 4

f (x) \geq 2

Solution steps

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

Domain of

x^{2} + 4 : - \infty < x < \infty

Extreme Points of

x^{2} + 4 :

Minimum

(0, 2)

Find the range for the interval

- \infty < x < \infty : 2 \leq f (x) < \infty

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

f (x) \geq 2

(0, - 1), (8, 7)

y = \frac{x - 1}{x ^{2} - 9}

ln (x + 6)

x + 3

\frac{0.052 x}{0.9 + 0.048 x}