Question

f(x)=\

$$\begin{cases}-\\dfrac{1}{4}x - 2, & x \\leq 2 \\\\ 2x + 1, & x > 2\\end{cases}$$

evaluate f(-1)

Explanation:

Step1: Determine the applicable function

Since \(-1 \leq 2\), we use the first part of the piece - wise function \(f(x)=-\frac{1}{4}x - 2\).

Step2: Substitute \(x = - 1\) into the function

Substitute \(x=-1\) into \(f(x)=-\frac{1}{4}x - 2\), we get \(f(-1)=-\frac{1}{4}\times(-1)-2\).
First, calculate \(-\frac{1}{4}\times(-1)=\frac{1}{4}\). Then, \(f(-1)=\frac{1}{4}-2\).
To subtract, we rewrite \(2\) as \(\frac{8}{4}\), so \(\frac{1}{4}-\frac{8}{4}=\frac{1 - 8}{4}=-\frac{7}{4}\).

Answer:

\(-\frac{7}{4}\)