Question

Evaluate the function as indicated. Determine its domain and range. \(f(x)=\left\\{\begin{array}{l}|x|+1, x<1 \\ -x+1, x \geq 1\end{array}\right.\) (a) \(f(-3)\) (b) \(f(1)\) (c) \(f(3)\) (d) \(f\left(b^{2}+1\right)\)

Short answer

The evaluated values are: (a) \(f(-3)=4\), (b) \(f(1)=0\), (c) \(f(3)=-2\), (d) \(f\left(b^{2}+1\right)\) depends on \(b\). The domain of function \(f(x)\) is all real numbers and its range is \(-\infty \leq y \leq 4\).

Step-by-step solution

Evaluate the function at specific points

To evaluate the function at a specific value, check which part of the function applies for that value of \(x\) and then substitute the value in the correct part. (a) To evaluate \(f(-3)\), since -3 is less than 1, we use the first part of the function \(-3 |+1| = |-3|+1 = 4\).(b) To evaluate \(f(1)\), since 1 is equal to 1, we use the second part of the function \(1 - 1 = 0\).(c) To evaluate \(f(3)\), since 3 is greater than 1, we use the second part of the function \(1 - 3 = -2\).(d) To find \(f\left(b^{2}+1\right)\), depending on the value of \(b\), either part of the function could apply. If \(b^{2}+1 < 1\), use the first part; otherwise, use the second part.

Determine the domain of the function

The domain is the set of all possible inputs, or x-values. For this function, there is no restriction on \(x\), it can be any real number. Thus, the domain of \(f(x)\) is all real numbers, or \(-\infty\) to \(+\infty\).

Determine the range of the function

The range is the set of all possible outputs, or y-values. The output of the first part of the function \(|x|+1\), for \(x<1\), can be any value > 1, since the absolute value function gives non-negative output. The second part of the function \(-x + 1\), for \(x \geq 1\), produces non-positive outputs due to the negative sign in front of \(x\). Plus 1 shifts the outputs negatively by 1. Thus the range is from \(-\infty\) to \(+4\) inclusive.