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

The range of sqrt(1+x^2) is f(x)>= 1

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range of sqrt(1+x^2)

range

1 + x^{2}

f (x) \geq 1

Solution steps

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

Domain of

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

Extreme Points of

1 + x^{2} :

Minimum

(0, 1)

Find the range for the interval

- \infty < x < \infty : 1 \leq f (x) < \infty

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

f (x) \geq 1

f (x) = 5 (x - 4)^{3}

f (x) = e^{\frac{x}{x - 1}}

f (x) = \frac{1}{2} csc (3 x)

3 x - 4

f (x) = \frac{1}{x} + x e^{- x}