The value of b, for which the function \(f(x)=\begin{cases}5x-4, & 0\lt x\le 1\\ 4x^2+3bx, & 1\lt x\lt 2\end{cases}\) is continuous at every point of its domain is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → −1
Each piece is a polynomial, so f is continuous inside each interval. The only point to check is the joining point x = 1.
\(f(1) = 5(1)-4 = 1\)
\( \displaystyle \lim_{x\to 1^-}f(x) = 5(1)-4 = 1 \)
\( \displaystyle \lim_{x\to 1^+}f(x) = 4(1)^2+3b(1) = 4+3b \)
For continuity at x = 1 the right-hand limit must equal f(1):
\(4+3b = 1\)
\(b = -1\)