Question

f(x)=\

$$\begin{cases}|2x + 6|, & x \\leq -3 \\\\ 2x^2, & -3 < x \\leq 0 \\\\ 5, & x > 0\\end{cases}$$

evaluate f(-1)

Explanation:

Step1: Determine the domain of -1

We check which piece of the piece - wise function \(x = - 1\) belongs to. The intervals are \(x\leq - 3\), \(-3\lt x\leq0\), and \(x > 0\). Since \(-3\lt - 1\leq0\), we use the function \(f(x)=2x^{2}\) for \(x=-1\).

Step2: Substitute x = -1 into the function

Substitute \(x=-1\) into \(f(x) = 2x^{2}\). We know that for any real number \(a\), \(a^{2}=(-a)^{2}\), so \((-1)^{2}=1\). Then \(f(-1)=2\times(-1)^{2}\).
Calculate \(2\times(-1)^{2}=2\times1 = 2\).

Answer:

2