Question

Evaluate the function at the indicated values if possible. If an indicated value is not in the​ domain, say so.

Answer 1

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:

$f(x)=\frac{x+7}{x^2-9}$ is the given function.

We are asked to find:

$f(-7)$

$f(2)$

$f(3)$.

f(-7) means to replace x in the expression called f with -7:

Evaluate $\frac{x+7}{x^2-9}$ at $x=-7$

$\frac{(-7)+7}{(-7)^2-9}$

$\frac{0}{49-9}$

$\frac{0}{40}$

$0$

So f(-7)=0.

f(2) means to replace x in the expression called f with 2:

Evaluate $\frac{x+7}{x^2-9}$ at $x=2$

$\frac{2+7}{2^2-9}$

$\frac{9}{4-9}$

$\frac{9}{-5}$

$\frac{-9}{5}$

So f(2)=-9/5

f(3) means to replace x in the expression called f with 3:

Evaluate $\frac{x+7}{x^2-9}$ at $x=3$

$\frac{3+7}{3^2-9}$

$\frac{10}{9-9}$

$\frac{10}{0}$

Division by 0 is not allowed so 3 is not in the domain of our function.