Question

3.
$f(x)=\

$$\begin{cases}\\ \\dfrac{3}{x + 4},&x < -5\\\\x^2 - 3x,&-5 < x \\leq 0\\\\x^4 - 7,&x > 0\\end{cases}$$

$
a. $f(-1) =$
b. $f(4) =$
c. $f(-10) =$
d. $f(0) =$

Explanation:

Step1: Identify rule for $f(-1)$

$-5 < -1 \leq 0$, use $f(x)=x^2-3x$

Step2: Substitute $x=-1$

$f(-1)=(-1)^2 - 3(-1) = 1 + 3 = 4$

Step3: Identify rule for $f(4)$

$4 > 0$, use $f(x)=x^4-7$

Step4: Substitute $x=4$

$f(4)=4^4 - 7 = 256 - 7 = 249$

Step5: Identify rule for $f(-10)$

$-10 < -5$, use $f(x)=\frac{3}{x+4}$

Step6: Substitute $x=-10$

$f(-10)=\frac{3}{-10+4}=\frac{3}{-6}=-\frac{1}{2}$

Step7: Identify rule for $f(0)$

$-5 < 0 \leq 0$, use $f(x)=x^2-3x$

Step8: Substitute $x=0$

$f(0)=0^2 - 3(0) = 0 - 0 = 0$

Answer:

a. $4$
b. $249$
c. $-\frac{1}{2}$
d. $0$