Question

1. two piecewise functions are shown below:

$f(x)= \

$$\begin{cases} 3x - 4 & \\text{for } x \\leq 5 \\\\ 2 - x & \\text{for } x > 5 \\end{cases}$$

$
$g(x)= \

$$\begin{cases} x^2 - 1 & \\text{for } x < 3 \\\\ 5 & \\text{for } x \\geq 3 \\end{cases}$$

$
what is the value of $4f(5) + 2g(3)$?
a 10
b 17
c 54
d 60

1. a function is shown below.

$h(x) = \

$$\begin{cases} -\\frac{1}{2}x - 15 & \\text{for } x \\leq -4 \\\\ 20 - 3x^2 & \\text{for } x > -4 \\end{cases}$$

$
what is the value of $h(-4) + 3h(-2)$?

Explanation:

Question 9

Step1: Find \( f(5) \)

Since \( 5 \leq 5 \), we use \( f(x) = 3x - 4 \). Substitute \( x = 5 \):
\( f(5) = 3(5) - 4 = 15 - 4 = 11 \)

Step2: Find \( g(3) \)

Since \( 3 \geq 3 \), we use \( g(x) = 5 \). So \( g(3) = 5 \)

Step3: Calculate \( 4f(5) + 2g(3) \)

Substitute \( f(5) = 11 \) and \( g(3) = 5 \):
\( 4(11) + 2(5) = 44 + 10 = 54 \)

Step1: Find \( h(-4) \)

Since \( -4 \leq -4 \), we use \( h(x) = -\frac{1}{2}x - 15 \). Substitute \( x = -4 \):
\( h(-4) = -\frac{1}{2}(-4) - 15 = 2 - 15 = -13 \)

Step2: Find \( h(-2) \)

Since \( -2 > -4 \), we use \( h(x) = 20 - 3x^2 \). Substitute \( x = -2 \):
\( h(-2) = 20 - 3(-2)^2 = 20 - 3(4) = 20 - 12 = 8 \)

Step3: Calculate \( h(-4) + 3h(-2) \)

Substitute \( h(-4) = -13 \) and \( h(-2) = 8 \):
\( -13 + 3(8) = -13 + 24 = 11 \)

Answer:

C. 54

Question 10