Question

additional exercises 6.1
form ii
rational expressions and functions: multiplying and dividing

find the function value. if a function value does not exist, so state.

1. $f(x)=\frac{x^{2}-3x-4}{x+1}$; $f(-3)$, $f(-1)$, $f(4)$

Explanation:

Step1: Substitute $x=-3$ into $f(x)$

$f(-3)=\frac{(-3)^2 - 3(-3) - 4}{-3 + 1}$

Step2: Simplify numerator and denominator

$f(-3)=\frac{9 + 9 - 4}{-2}=\frac{14}{-2}=-7$

Step3: Substitute $x=-1$ into $f(x)$

$f(-1)=\frac{(-1)^2 - 3(-1) - 4}{-1 + 1}$

Step4: Check denominator validity

Denominator $=-1+1=0$, so $f(-1)$ is undefined (function value does not exist, since division by zero is not allowed).

Step5: Substitute $x=4$ into $f(x)$

$f(4)=\frac{(4)^2 - 3(4) - 4}{4 + 1}$

Step6: Simplify numerator and denominator

$f(4)=\frac{16 - 12 - 4}{5}=\frac{0}{5}=0$

Answer:

$f(-3)=-7$, $f(-1)$ does not exist, $f(4)=0$