Question

For what value of the constant c is the function f continuous on \(\left( { - \infty ,\infty } \right)\)?

47. \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{c{x^{\bf{2}}} + {\bf{2}}x}&{{\bf{if}}\,\,\,x < {\bf{2}}}\\{{x^3} - cx}&{{\bf{if}}\,\,\,x \ge {\bf{2}}}\end{array}} \right.\)

Short answer

The value is \(c = \frac{2}{3}\).

Step-by-step solution

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

Take left-hand limit;

\(\begin{aligned}\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) &= \mathop {\lim }\limits_{x \to {2^ - }} c{x^2} + 2x\\ &= c{\left( 2 \right)^2} + 2\left( 2 \right)\\ &= 4\left( {c + 1} \right)\end{aligned}\)

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

Take right-hand limit;

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

Find the value of c

As the limit has to exist, so, the left andright-hand limitmust be equal.

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

So, the value of c should be \(\frac{2}{3}\).