Answer:
f(-7)=0.
f(2)=-9/5.
f(3) doesn't exist because 3 isn't in the domain of the function.
Step-by-step explanation:
[tex]f(x)=\frac{x+7}{x^2-9}[/tex] is the given function.
We are asked to find:
[tex]f(-7)[/tex]
[tex]f(2)[/tex]
[tex]f(3)[/tex].
f(-7) means to replace x in the expression called f with -7:
Evaluate [tex]\frac{x+7}{x^2-9}[/tex] at [tex]x=-7[/tex]
[tex]\frac{(-7)+7}{(-7)^2-9}[/tex]
[tex]\frac{0}{49-9}[/tex]
[tex]\frac{0}{40}[/tex]
[tex]0[/tex]
So f(-7)=0.
f(2) means to replace x in the expression called f with 2:
Evaluate [tex]\frac{x+7}{x^2-9}[/tex] at [tex]x=2[/tex]
[tex]\frac{2+7}{2^2-9}[/tex]
[tex]\frac{9}{4-9}[/tex]
[tex]\frac{9}{-5}[/tex]
[tex]\frac{-9}{5}[/tex]
So f(2)=-9/5
f(3) means to replace x in the expression called f with 3:
Evaluate [tex]\frac{x+7}{x^2-9}[/tex] at [tex]x=3[/tex]
[tex]\frac{3+7}{3^2-9}[/tex]
[tex]\frac{10}{9-9}[/tex]
[tex]\frac{10}{0}[/tex]
Division by 0 is not allowed so 3 is not in the domain of our function.
