Question

1. ( f(x) = \begin{cases} x^2 + 2, & x leq 1 \\ 2x^2 + 2, & x > 1 end{cases}

a) ( f(-2)
b) ( f(1)
c) ( f(2) )

Explanation:

Part (a): \( f(-2) \)

Step1: Determine the applicable function

Since \(-2 \leq 1\), we use \( f(x) = x^2 + 2 \).

Step2: Substitute \( x = -2 \)

$$ f(-2) = (-2)^2 + 2 $$

Step3: Calculate the value

$$ (-2)^2 = 4 \implies f(-2) = 4 + 2 = 6 $$

Step1: Determine the applicable function

Since \( 1 \leq 1 \), we use \( f(x) = x^2 + 2 \).

Step2: Substitute \( x = 1 \)

$$ f(1) = 1^2 + 2 $$

Step3: Calculate the value

$$ 1^2 = 1 \implies f(1) = 1 + 2 = 3 $$

Step1: Determine the applicable function

Since \( 2 > 1 \), we use \( f(x) = 2x^2 + 2 \).

Step2: Substitute \( x = 2 \)

$$ f(2) = 2(2)^2 + 2 $$

Step3: Calculate the value

$$ 2^2 = 4 \implies 2(4) = 8 \implies f(2) = 8 + 2 = 10 $$

Answer:

\( 6 \)

Part (b): \( f(1) \)