Question

Find the values of a and b that make f continuous everywhere.

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{{{x^{\bf{2}}} - {\bf{4}}}}{{{\bf{x}} - {\bf{2}}}}}&{{\bf{if}}\,\,\,x < {\bf{2}}}\\{a{x^{\bf{2}}} - bx + {\bf{3}}}&{{\bf{if}}\,\,\,{\bf{2}} \le x < {\bf{3}}}\\{{\bf{2}}x - a + b}&{{\bf{if}}\,\,\,x \ge {\bf{3}}}\end{array}} \right.\)

Short answer

The values are \(a = \frac{1}{2}\), and \(b = \frac{1}{2}\).

Step-by-step solution

Check the limits of \(f\left( x \right)\) at \(x = {\bf{2}}\)

Find theleft-hand limitfor \(f\left( x \right)\) at \(x = 2\).

\(\begin{aligned}\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) &= \mathop {\lim }\limits_{x \to {2^ - }} \left( {\frac{{{x^2} - 4}}{{x - 2}}} \right)\\ &= \mathop {\lim }\limits_{x \to {2^ - }} \left( {\frac{{\left( {x - 2} \right)\left( {x + 4} \right)}}{{x - 2}}} \right)\\ &= \mathop {\lim }\limits_{x \to {2^ - }} \left( {x + 4} \right)\\ &= 4\end{aligned}\)

Find theright-hand limitfor \(f\left( x \right)\) at \(x = 2\).

\(\begin{aligned}\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) &= \mathop {\lim }\limits_{x \to {2^ + }} \left( {a{x^2} - bx + 3} \right)\\ &= a\left( {{2^2}} \right) - b\left( 2 \right) + 3\\ &= 4a - 2b + 3\end{aligned}\)

As f is continuous everywhere, therefore left and right-hand limit must be equal.

\(\begin{array}{c}4a - 2b + 3 = 4\\4a - 2b = 1\end{array}\)

Check the limits of \(f\left( x \right)\) at \(x = {\bf{3}}\)

Find the left-hand limit for \(f\left( x \right)\) at \(x = 3\).

\(\begin{aligned}\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) &= \mathop {\lim }\limits_{x \to {3^ - }} \left( {a{x^2} - bx + 3} \right)\\ &= a\left( {{3^2}} \right) - b\left( 3 \right) + 3\\ &= 9a - 3b + 3\end{aligned}\)

Find the right-hand limit for \(f\left( x \right)\) at \(x = 3\).

\(\begin{aligned}\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) &= \mathop {\lim }\limits_{x \to {3^ + }} \left( {2x - a + b} \right)\\ &= 6 - a + b\end{aligned}\)

As f is continuous everywhere, therefore, left and right-hand limits must be equal.

\(\begin{array}{c}9a - 3b + 3 = 6 - a + b\\10a - 4b = 3\end{array}\)

Find the values of a and b

Solve the equations \(4a - 2b = 1\) and \(10a - 4b = 3\).

\(\begin{aligned}10a - 2\left( {4a - 1} \right) &= 3\\10a - 8a + 2 &= 3\\a &= \frac{1}{2}\end{aligned}\)

Then,

\(\begin{aligned}4\left( {\frac{1}{2}} \right) - 2b &= 1\\b &= \frac{1}{2}\end{aligned}\)

So, the values of a and b are equal to \(\frac{1}{2}\).