Chapter 4, Problem 76RE

Solution

To calculate: The domain and range for the given function.

Domain: $\left[-1,0\right]$

Range: $\left[-3,\pi -3\right]$

Given information: $y={\cos }^{-1}(2x+1)-3$

Calculation: First, note the domain and range of the parent function: $y={\sin }^{-1}x$ .

Domain: $[-1,1]$ and range: $\left[0,\pi \right]$

The function can be written as:

  $f(x)={\cos }^{-1}\left(\frac{1}{1/2}x+1\right)-3$

Which is a horizontal shrink by 1 unit to the left, horizontal shrink by a factor of $\frac{1}{2}$ , and vertical shift by 3 units down of the parent function.

The domain of $f(x)$ is the same as the parent function decreased by 1 because of the horizontal shift then multiplied by $\frac{1}{2}$ , because of the horizontal shrinks.

  $\left[\frac{1}{2}(-1-1),\frac{1}{2}(1-1)\right]$

Domain: $\left[-1,0\right]$

The range of $f(x)$ is the range of the parent function decreased by 3 because of the vertical shift:

Range: $\left[-3,\pi -3\right]$

From $y={\cos }^{-1}x$ : horizontal shrink shift by 1 unit to the right, horizontal shrink by a factor of $\frac{1}{3}$ and vertical shift by 2 units up;

Domain: $\left[-1,0\right]$ ;

Range: $\left[-3,\pi -3\right]$