To find the value of the constant [tex]\( c \)[/tex] for which the function [tex]\( f \)[/tex] is continuous on [tex]\((0, \infty)\)[/tex], we need to ensure that the two parts of the function meet smoothly at [tex]\( x = 2 \)[/tex]. This involves ensuring that the left-hand limit of the function as [tex]\( x \)[/tex] approaches 2 is equal to the right-hand value of the function at [tex]\( x = 2 \)[/tex].
Step-by-step solution:
1. Determine the left-hand limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 2:
For [tex]\( 0 \leq x < 2 \)[/tex], the function is defined as [tex]\( f(x) = 2x + c \)[/tex].
The left-hand limit (as [tex]\( x \)[/tex] approaches 2 from the left, denoted as [tex]\( x \to 2^- \)[/tex]):
[tex]\[
\lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^-}} (2x + c) = 2(2) + c = 4 + c
\][/tex]
2. Determine the value of [tex]\( f(x) \)[/tex] at [tex]\( x = 2 \)[/tex] from the right-hand side:
For [tex]\( x \geq 2 \)[/tex], the function is defined as [tex]\( f(x) = c x^2 - 2 \)[/tex].
Evaluate this function at [tex]\( x = 2 \)[/tex]:
[tex]\[
f(2) = c(2)^2 - 2 = 4c - 2
\][/tex]
3. Set the left-hand limit equal to the value at [tex]\( x = 2 \)[/tex]:
To ensure continuity at [tex]\( x = 2 \)[/tex], these two expressions must be equal:
[tex]\[
4 + c = 4c - 2
\][/tex]
4. Solve for [tex]\( c \)[/tex]:
[tex]\[
4 + c = 4c - 2
\][/tex]
Subtract [tex]\( c \)[/tex] from both sides to gather all [tex]\( c \)[/tex]-terms on one side:
[tex]\[
4 = 3c - 2
\][/tex]
Add 2 to both sides:
[tex]\[
6 = 3c
\][/tex]
Divide by 3:
[tex]\[
c = 2
\][/tex]
Therefore, the function [tex]\( f \)[/tex] is continuous on [tex]\((0, \infty)\)[/tex] when the constant [tex]\( c \)[/tex] is [tex]\( \boxed{2} \)[/tex].
