Question

The function f(x) is defined as follows: 

\[f\left( x \right) = \begin{cases}x^2 + ax + b , & 0 \leq x < 2 \\ 3x + 2 , & 2 \leq x \leq 4 \\ 2ax + 5b , & 4 < x \leq 8\end{cases}\]

If f is continuous on [0, 8], find the values of a and b.

Answer 1

Given: f is continuous on  \[\left[ 0, 8 \right]\] . 

∴ f is continuous at x = 2 and x = 4

At x = 2, we have

\[\lim_{x \to 2^-} f\left( x \right) = \lim_{h \to 0} f\left( 2 - h \right) = \lim_{h \to 0} \left[ \left( 2 - h \right)^2 + a\left( 2 - h \right) + b \right] = 4 + 2a + b\]

\[\lim_{x \to 2^+} f\left( x \right) = \lim_{h \to 0} f\left( 2 + h \right) = \lim_{h \to 0} \left[ 3\left( 2 + h \right) + 2 \right] = 8\]

Also,
At x = 4, we have

\[\lim_{x \to 4^-} f\left( x \right) = \lim_{h \to 0} f\left( 4 - h \right) = \lim_{h \to 0} \left[ 3\left( 4 - h \right) + 2 \right] = 14\]

\[\lim_{x \to 4^+} f\left( x \right) = \lim_{h \to 0} f\left( 4 + h \right) = \lim_{h \to 0} \left[ 2a\left( 4 + h \right) + 5b \right] = 8a + 5b\]

f is continuous at x = 2 and x = 4

∴  \[\lim_{x \to 2^-} f\left( x \right) = \lim_{x \to 2^+} f\left( x \right) and \lim_{x \to 4^-} f\left( x \right) = \lim_{x \to 4^+} f\left( x \right)\]

\[\Rightarrow 4 + 2a + b = 8\text{  and } 8a + 5b = 14\]
\[ \Rightarrow 2a + b = 4 . . . \left( 1 \right) \text{ and }  8a + 5b = 14 . . . \left( 2 \right)\]

On simplifying eqs. (1) and (2), we get

\[a = 3 \text{ and }  b = - 2\]