What is the range of h(x)=sqrt(x+1) ?

The range of h(x)=sqrt(x+1) is f(x)>= 0

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range of h(x)=sqrt(x+1)

range

h (x) = x + 1

f (x) \geq 0

Interval Notation

[0, \infty)

Solution steps

The range of an radical function of the form

c a x + b + k

f (x) \geq k

k = 0

f (x) \geq 0

f (x) = e^{x - 1} + 2

f (x) = \frac{1}{1 + x ^{2}}

f (x) = 1 - 5 x

f (x) = \frac{2}{x ^{2} - 9}

f (x) = \frac{2 ( x + 2 )}{x ^{2} - x - 6}