The value of [tex]g(-4)=\frac{-4}{3}, \ g(-2)=3 \ \text{and} \ g(2)=\frac{8}{3}[/tex] .
Solution:
Given function:
[tex]g(x)=\left\{\begin{array}{ll}-\frac{1}{3} x^{2}+4 & \text { if } x \neq-2 \\3 & \text { if } x=-2\end{array}\right.[/tex]
Substitute x = –4 in g(x), we get
[tex]$g(-4) =- \frac{1}{3}(-4)^2+4[/tex]
[tex]$=- \frac{16}{3}+4[/tex]
To make the denominator same, multiply and divide the 2nd term by 3.
[tex]$=- \frac{16}{3}+\frac{12}{3}[/tex]
[tex]$g(-4)=- \frac{4}{3}[/tex]
It is given that, if x = –2, then g(x) = 3
Therefore g(–2) = 3
Substitute x = 2 in g(x), we get
[tex]$g(2) =- \frac{1}{3}(2)^2+4[/tex]
[tex]$=- \frac{4}{3}+4[/tex]
To make the denominator same, multiply and divide the 2nd term by 3.
[tex]$=- \frac{4}{3}+\frac{12}{3}[/tex]
[tex]$g(-2)= \frac{8}{3}[/tex]
The value of [tex]g(-4)=\frac{-4}{3}, \ g(-2)=3 \ \text{and} \ g(2)=\frac{8}{3}[/tex] .
