Question

1. two piecewise functions are shown below.

$h(x) = \

$$\begin{cases} -3x & \\text{for } x < 2 \\\\ 4x + 1 & \\text{for } x \\geq 2 \\end{cases}$$

$
$g(x) = \

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

$
what is the value of $3h(2) + 4g(1)$?
a 39
b 28
c 10
d -6

1. given $f(x) = 2x^2 - 3x + 1$ and $g(x) = 4x - 5$, find $f(2) + g(2)$.

a. 4
b. 6
c. 11
d. 24

Explanation:

Problem 7

Step1: Find \( h(2) \)

For \( h(x) \), when \( x = 2 \), we use the rule \( 4x + 1 \) (since \( x \geq 2 \)).
\( h(2)=4(2)+1 = 8 + 1 = 9 \)

Step2: Find \( g(1) \)

For \( g(x) \), when \( x = 1 \), we use the rule \( x^2 + 2 \) (since \( x < 3 \)).
\( g(1)=1^2 + 2 = 1 + 2 = 3 \)

Step3: Calculate \( 3h(2)+4g(1) \)

Substitute \( h(2) = 9 \) and \( g(1) = 3 \) into the expression.
\( 3(9)+4(3)=27 + 12 = 39 \)

Step1: Find \( f(2) \)

Substitute \( x = 2 \) into \( f(x)=2x^2 - 3x + 1 \).
\( f(2)=2(2)^2-3(2)+1 = 2(4)-6 + 1 = 8 - 6 + 1 = 3 \)

Step2: Find \( g(2) \)

Substitute \( x = 2 \) into \( g(x)=4x - 5 \).
\( g(2)=4(2)-5 = 8 - 5 = 3 \)

Step3: Calculate \( f(2)+g(2) \)

Add the two results: \( 3 + 3 = 6 \)

Answer:

A. 39

Problem 8