Question

Find the values of a so that the function 

\[f\left( x \right) = \begin{cases}ax + 5, if & x \leq 2 \\ x - 1 , if & x > 2\end{cases}\text{is continuous at x} = 2 .\]

Answer 1

Given:

\[f\left( x \right) = \binom{ax + 5, if x \leq 2}{x - 1, if x > 2}\]

 We observe

(LHL at x = 2) = 

\[\lim_{h \to 0} a\left( 2 - h \right) + 5 = 2a + 5\]

(RHL at x = 2) = \[\lim_{x \to 2^+} f\left( x \right) = \lim_{h \to 0} f\left( 2 + h \right)\]

\[\lim_{h \to 0} \left( 2 + h - 1 \right)\]

\[1\]

And,

\[f\left( 2 \right) = a\left( 2 \right) + 5 = 2a + 5\]

Since

\[f\left( x \right)\]  is continuous at x = 2, we have

\[\lim_{x \to 2^-} f\left( x \right) = \lim_{x \to 2^+} f\left( x \right) = f\left( 2 \right)\]

\[2a + 5 = 1\]

\[2a = - 4\]

\[a = - 2\]