For what value of the constant $c$ is the function $$f\left(x\right)=\left\{\begin{array}{ll}2x+1& \text{, }x<3\\ c& \text{, }x=3\\ x^2-2& \text{, }x>3\end{array}\right\}$$ continuous on $ℝ$?

• A. $c=5$
• B. $c=4$
• C. $c=6$
• D. $c=7$

Answer: (C)

Verified step by step guidance

1. Step 1: Recall the definition of continuity. A function f(x) is continuous at a point x = a if the following three conditions are satisfied: (1) f(a) is defined, (2) the limit of f(x) as x approaches a exists, and (3) the limit of f(x) as x approaches a is equal to f(a).
2. Step 2: Analyze the given piecewise function. The function f(x) is defined as f(x) = 2x + 1 for x < 3 and f(x) = c for x ≥ 3. To ensure continuity at x = 3, the left-hand limit (as x approaches 3 from the left) must equal the right-hand limit (as x approaches 3 from the right), and both must equal f(3).
3. Step 3: Compute the left-hand limit as x approaches 3 from the left. For x < 3, f(x) = 2x + 1. Thus, the left-hand limit is lim_{x → 3^-} f(x) = lim_{x → 3^-} (2x + 1).
4. Step 4: Compute the right-hand limit as x approaches 3 from the right. For x ≥ 3, f(x) = c. Thus, the right-hand limit is lim_{x → 3^+} f(x) = c.
5. Step 5: Set the left-hand limit equal to the right-hand limit to ensure continuity. Solve the equation lim_{x → 3^-} (2x + 1) = c. Substitute x = 3 into the left-hand limit expression to find c. This will give the value of c that makes the function continuous on ℝ.