Question

43-45 Find the numbers at which f is discontinuous. At which of these numbers is f continuous from the right, from the left, or neither? Sketch the graph of f?

45. \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{x + {\bf{2}}}&{{\bf{if}}\,\,\,x < {\bf{0}}}\\{{e^x}}&{{\bf{if}}\,\,\,{\bf{0}} \le x \le {\bf{1}}}\\{{\bf{2}} - x}&{{\bf{if}}\,\,\,x > {\bf{1}}}\end{array}} \right.\)

Short answer

At \(x = 0\) from the right and \(x = 1\) from the left.

Step-by-step solution

Sketch the graph of f

The figure below represents the graph of \(f\left( x \right)\).

Check the function \(f\left( x \right)\) 

The function \(f\left( x \right)\) is continuous in the interval \(\left( { - \infty ,0} \right)\), \(\left( {0,1} \right)\), and \(\left( {1,\infty } \right)\).

Find the limit of the function at \(x = {\bf{0}}\)

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

\(\begin{aligned}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) &= \mathop {\lim }\limits_{x \to {0^ - }} \left( {x + 2} \right)\\ &= 0 + 2\\ &= 2\end{aligned}\)

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

\(\begin{aligned} \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) &= \mathop {\lim }\limits_{x \to {0^ + }} {e^x}\\ &= {e^0}\\ &= 1\end{aligned}\)

The left-hand limit and right-hand limit are not equal at \(x = 0\). So, the function is not continuous at in \(x = 0\).

But \(f\left( 0 \right) = 1\), so f is continuous at \(x = 0\) from the right.

Find the limit of the function at \(x = {\bf{1}}\)

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

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

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

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

As left-hand limit and right-hand limit are not equal at \(x = 1\). So, the function is not continuous at in \(x = 1\).

But \(f\left( 1 \right) = e\), so f is continuous at \(x = 1\) from left.