Chapter 8, Problem 29RE

To determine

To find Range:

For this first we find inverse of h.
Squaring $h(x)=\sqrt{x+1}$

We get $\begin{array}{l}h{(}^x=x+1\\ \\ \end{array}$

  ${\{h(x)\}}^2-1=x$

Hence for any value of h(x) we get x as real.

But by definition of h(x) as positive square root of x+1

  $h(x)\ge 0$

Range = [0,infinity)

Conclusion:

For the given function h(x) domain and range are as follows:

Domain: [-1,infinity)

Range = [0,infinity)

1. To find:

Domain and range of the function k(x)

To determine

To Find:

Domain

The function k(x) has real values for all value of x greater than or equal to 0.

Hence domain = : [0,infinity)

To find Range:

For this first we find inverse of k.

Add 1 to both the sides and then
Taking square root for $x=\sqrt{k(x)+1}$

We get for real x, k(x) +1 must be positive.
i.e. k(x) $\ge -1$

Range = [-1,infinity)