Question

if the function
f(x)=\begin{cases}2ax^{2}+bx - 1&\text{if }xleq3\bx^{2}+bx - a&\text{if }x > 3end{cases}
is continuous for all real numbers (x), then

(19a - 15b=1)
(18a - 9b=1)
(19a + 15b=1)
(19a - 9b=1)

Explanation:

Step1: Recall continuity condition

For a function to be continuous at \(x = 3\), \(\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{+}}f(x)\).

Step2: Calculate left - hand limit

\(\lim_{x
ightarrow3^{-}}f(x)=2a(3)^{2}+b(3)-1 = 18a + 3b-1\).

Step3: Calculate right - hand limit

\(\lim_{x
ightarrow3^{+}}f(x)=b(3)^{2}+b(3)-a=9b + 3b-a= - a+12b\).

Step4: Set left - hand and right - hand limits equal

\(18a + 3b-1=-a + 12b\).

Step5: Rearrange the equation

\(18a+a+3b - 12b=1\), which simplifies to \(19a-9b = 1\).

Answer:

\(19a - 9b = 1\)