Suppose that the function [tex]f[/tex] is defined for all real numbers as follows:
[tex]\[ f(x)=\left\{\begin{array}{ll}
-\frac{1}{2} x^2+5 & \text{if } x \neq 2 \\
1 & \text{if } x=2
\end{array}\right. \][/tex]

Find [tex]f(-1)[/tex], [tex]f(2)[/tex], and [tex]f(5)[/tex].



Answer :

Let's evaluate the given piecewise function [tex]\( f(x) = \left\{\begin{array}{ll} -\frac{1}{2} x^2 + 5 & \text{if } x \neq 2 \\ 1 & \text{if } x = 2 \end{array}\right. \)[/tex] at the points [tex]\( x = -1 \)[/tex], [tex]\( x = 2 \)[/tex], and [tex]\( x = 5 \)[/tex].

### 1. Finding [tex]\( f(-1) \)[/tex]:
Since [tex]\( -1 \neq 2 \)[/tex], we use the formula [tex]\( f(x) = -\frac{1}{2} x^2 + 5 \)[/tex].

Substitute [tex]\( x = -1 \)[/tex] into the formula:
[tex]\[ f(-1) = -\frac{1}{2} (-1)^2 + 5 \][/tex]

Calculate [tex]\((-1)^2\)[/tex]:
[tex]\[ (-1)^2 = 1 \][/tex]

Multiply by [tex]\(-\frac{1}{2}\)[/tex]:
[tex]\[ -\frac{1}{2} \cdot 1 = -\frac{1}{2} \][/tex]

Add 5:
[tex]\[ -\frac{1}{2} + 5 = 4.5 \][/tex]

So, [tex]\( f(-1) = 4.5 \)[/tex].

### 2. Finding [tex]\( f(2) \)[/tex]:
Since [tex]\( 2 = 2 \)[/tex], we use the specific value given by the piecewise function for [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 1 \][/tex]

### 3. Finding [tex]\( f(5) \)[/tex]:
Since [tex]\( 5 \neq 2 \)[/tex], we use the formula [tex]\( f(x) = -\frac{1}{2} x^2 + 5 \)[/tex].

Substitute [tex]\( x = 5 \)[/tex] into the formula:
[tex]\[ f(5) = -\frac{1}{2} (5)^2 + 5 \][/tex]

Calculate [tex]\( (5)^2 \)[/tex]:
[tex]\[ (5)^2 = 25 \][/tex]

Multiply by [tex]\(-\frac{1}{2}\)[/tex]:
[tex]\[ -\frac{1}{2} \cdot 25 = -12.5 \][/tex]

Add 5:
[tex]\[ -12.5 + 5 = -7.5 \][/tex]

So, [tex]\( f(5) = -7.5 \)[/tex].

### Summary of Results:
[tex]\[ f(-1) = 4.5 \][/tex]
[tex]\[ f(2) = 1 \][/tex]
[tex]\[ f(5) = -7.5 \][/tex]

Thus, we have found that [tex]\( f(-1) = 4.5 \)[/tex], [tex]\( f(2) = 1 \)[/tex], and [tex]\( f(5) = -7.5 \)[/tex].