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. $f(x)=\frac{x^2 - 3x - 4}{x + 1}$; $f(-3)$, $f(-1)$, $f(4)$
2. does not exist
3. $f(x)=\frac{x^2 + 2x + 1}{2 - x}$; $f(-7)$, $f(-2)$
4. _______________

Explanation:

Step1: Solve $f(-3)$ for problem 1

Substitute $x=-3$ into $f(x)=\frac{x^2-3x-4}{x+1}$
$$f(-3)=\frac{(-3)^2-3(-3)-4}{-3+1}=\frac{9+9-4}{-2}=\frac{14}{-2}=-7$$

Step2: Solve $f(-1)$ for problem 1

Substitute $x=-1$ into $f(x)=\frac{x^2-3x-4}{x+1}$
$$f(-1)=\frac{(-1)^2-3(-1)-4}{-1+1}=\frac{1+3-4}{0}=\frac{0}{0}$$
Denominator is 0, so $f(-1)$ does not exist.

Step3: Solve $f(4)$ for problem 1

Substitute $x=4$ into $f(x)=\frac{x^2-3x-4}{x+1}$
$$f(4)=\frac{(4)^2-3(4)-4}{4+1}=\frac{16-12-4}{5}=\frac{0}{5}=0$$

Step4: Solve $f(-7)$ for problem 2

Substitute $x=-7$ into $f(x)=\frac{x^2+2x+1}{2-x}$
$$f(-7)=\frac{(-7)^2+2(-7)+1}{2-(-7)}=\frac{49-14+1}{9}=\frac{36}{9}=4$$

Step5: Solve $f(-2)$ for problem 2

Substitute $x=-2$ into $f(x)=\frac{x^2+2x+1}{2-x}$
$$f(-2)=\frac{(-2)^2+2(-2)+1}{2-(-2)}=\frac{4-4+1}{4}=\frac{1}{4}$$

Answer:

1. $f(-3)=-7$, $f(-1)$ does not exist, $f(4)=0$
2. $f(-7)=4$, $f(-2)=\frac{1}{4}$