Question

Where is the function \(h\left( x \right) = \left| {x - 1} \right| + \left| {x + 2} \right|\) differentiable? Give a formula for \(h'\) and sketch the graphs of \(h\) and \(h'\).

Short answer

\(h\) is not differential at \(x = 1\,\,{\rm{and }}x = - 2\). The formula for \(h'\) is \(h'\left( x \right) = \left\{ \begin{array}{l} - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,x < - 2\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\, - 2 < x < 1\\2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,x > 1\end{array} \right.\).

The graphs are:

\(h\left( x \right) \Rightarrow \)

\(h'\left( x \right) \Rightarrow \)

Step-by-step solution

Differentiable functions

Any function can be taken as a differentiable function if and only if that function can show the property of differentiability as well as continuity with any given range for the variable.

Evaluating given function and calculating left-hand and right-hand derivatives:

Thegiven function is:

\(\begin{align}h\left( x \right) &= \left| {x - 1} \right| + \left| {x + 2} \right|\\h\left( x \right) &= \left\{ \begin{array}{l} - 2x - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,x \le - 2\\3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\, - 2 < x < 1\\2x + 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,x \ge 1\end{array} \right.\\h'\left( x \right) &= \left\{ \begin{array}{l} - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,x < - 2\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\, - 2 < x < 1\\2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,x > 1\end{array} \right.\end{align}\)

Therefore, theformula for \(h'\) is \(h'\left( x \right) = \left\{ \begin{array}{l} - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,x < - 2\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\, - 2 < x < 1\\2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,x > 1\end{array} \right.\).

Calculating the left-hand derivatives of the function at\(x = 1\)using equation 2 as:

\(\begin{align}h_ - ^{'}\left( 1 \right) &= \mathop {\lim }\limits_{x \to {1^ - }} \frac{{h\left( x \right) - h\left( 1 \right)}}{{x - 1}}\\ &= 0\end{align}\)

Again, calculating the right-hand derivatives of the function at\(x = 1\)using equation 2 as:

\(\begin{align}h_ + ^{'}\left( 1 \right) &= \mathop {\lim }\limits_{x \to {1^ + }} \frac{{h\left( x \right) - h\left( 1 \right)}}{{x - 1}}\\ &= 2\end{align}\)

Comparing results.

From the above calculation, we have:

\(h_ - ^{'}\left( 1 \right) \ne h_ + ^{'}\left( 1 \right)\)

Similarly,

\(h_ - ^{'}\left( { - 2} \right) \ne h_ + ^{'}\left( { - 2} \right)\)

Thus, the function \(h'\left( 1 \right)\,\,\& \,\,h'\left( { - 2} \right)\) does not exist.

Hence, the function \(h\) is not differential at \(x = 1\) and \(x = - 2\).

Plotting Graph.

The graph for \(h\) and \(h'\) can be drawn as shown:

\(h\left( x \right) \Rightarrow \)

\(h'\left( x \right) \Rightarrow \)

Hence, these are the required graphs.