Let $f\left(x\right)$ = $\left\{\begin{array}{cc}ax+b& \text{,}x<2\\ x^2& \text{,}x\ge 2\end{array}\right\}$. For which values of $a$ and $b$ is $f\left(x\right)$ continuous everywhere?

• A. $a=2$, $b=0$
• B. $a=1$, $b=2$
• C. $a=4$, $b=-4$
• D. $a=0$, $b=4$

Answer: (C)

Verified step by step guidance

1. Step 1: Understand the definition of continuity. A function is continuous everywhere if there are no breaks, jumps, or holes in its graph. This means the left-hand limit, right-hand limit, and the function value at a given point must all be equal.
2. Step 2: Analyze the given piecewise function f(x). For x < 2, f(x) = ax + b, and for x ≥ 2, f(x) = x². To ensure continuity at x = 2, the value of f(x) from both pieces must match at this point.
3. Step 3: Set up the condition for continuity at x = 2. The left-hand limit of f(x) as x approaches 2 from the left is f(2) = a(2) + b. The right-hand limit of f(x) as x approaches 2 from the right is f(2) = 2² = 4. Equate these two expressions: a(2) + b = 4.
4. Step 4: Solve for a and b using the equation derived in Step 3. Rearrange the equation to express b in terms of a: b = 4 - 2a. Substitute different values of a and b from the options provided to check which pair satisfies the equation.
5. Step 5: Verify the correct pair of values (a, b) that ensures continuity everywhere. Substitute the correct values into the equation and confirm that both the left-hand limit and right-hand limit at x = 2 are equal to the function value at x = 2.