Let g(x) equal 2x - 4 when -2 is less than x and x is at most 5, and equal x^2 - 2x + 3 when x is greater than 5. Evaluate g(-1), g(7), g(0) and g(-2).

Step-by-step solution

1. 1

   Adopt the right order of operations for a piecewise function. For each input, first decide which branch applies by testing the conditions, and only then substitute. Substituting into the wrong branch is the whole failure mode here.

2. 2

   (a) g(-1). Is $-2<-1\le 5$? Yes, so use the first branch:

   $g(-1)=2(-1)-4=-2-4=-6.$

3. 3

   (b) g(7). Is $7>5$? Yes, so use the second branch:

   $g(7)=7^{2}-2(7)+3=49-14+3=38.$

4. 4

   (c) g(0). Is $-2<0\le 5$? Yes, first branch again:

   $g(0)=2(0)-4=-4.$

5. 5

   (d) g(-2) — check the inequality symbols carefully. The first branch requires $-2<x$, a strict inequality, so $x=-2$ does not qualify. The second requires $x>5$, which $-2$ also fails. No branch covers $-2$, so

   $g(-2)\ \text{is undefined.}$

   Had the condition read $-2\le x\le 5$, the answer would have been $2(-2)-4=-8$. The single character is the whole question.

6. 6

   State the domain that follows. The two conditions together cover $(-2,5]\cup(5,\infty)=(-2,\infty)$, so $g$ is defined for every $x>-2$ and nowhere else. Note the pieces even join up: the first branch reaches $2(5)-4=6$ at $x=5$, and the second approaches $25-10+3=18$ just past it, so the graph jumps at $x=5$.