Question

(a)For what values of \(x\) is the function \(f\left( x \right) = \left| {{x^2} - 9} \right|\) differentiable? Find a formula for \(f'\).

(b)Sketch the graphs of \(f\) and \(f'\).

Short answer

1. \(f\)is not differential at \(x = \pm 3\). The formula for \(f'\left( x \right)\) is \(f'\left( x \right) = \left\{ \begin{array}{l}2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,\left| x \right| > 3\\ - 2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,\left| x \right| < 3\end{array} \right.\)
2. The graphs are:

\(f\left( x \right) \Rightarrow \)

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

Step-by-step solution

(a) Step 1: Differentiable functions

The differentiable functions are those functions which can be differentiable within the interval of convergence and they must be continuous functions for all values of variable.

Evaluating given function:

Thegiven function is:

\(\begin{align}f\left( x \right) &= \left| {{x^2} - 9} \right|\\&{x^2} - 9 < 0\,\,\,\,\,\,\,\,\,\,\,\,{\rm{for}}\,\,{x^2} < 9\\&\left| x \right| < 3\\& - 3 < x < 3\end{align}\)

Using this, we have:

\(\begin{align}f\left( x \right) &= \left\{ \begin{array}{x^2} - 9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,x \le - 3\\ - {x^2} + 9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\, - 3 < x < 3\\{x^2} - 9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,x \ge 3\end{array} \right.\\f'\left( x \right) &= \left\{ \begin{array}2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,\left| x \right| > 3\\ - 2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,\left| x \right| < 3\end{array} \right.\end{align}\)

Thus, the formula for \(f'\left( x \right)\) is \(f'\left( x \right) = \left\{ \begin{array}{l}2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,\left| x \right| > 3\\ - 2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,\left| x \right| < 3\end{array} \right.\).

Calculating left-hand and right-hand derivatives.

Calculating the left-hand derivatives of the function at\(x = 3\)using equation 2 as:

\(\begin{align}f_ - ^{'}\left( 3 \right) &= \mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left( {3 + h} \right) - f\left( 3 \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to {0^ - }} \frac{{\left( { - {{\left( {3 + h} \right)}^2} + 9} \right) - \left( 0 \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to {0^ - }} \frac{{ - h\left( {6 + h} \right)}}{h}\\ &= - 6\end{align}\)

Again, calculating the right-hand derivatives of the function at\(x = 3\)using equation 2 as:

\(\begin{align}f_ + ^{'}\left( 3 \right) &= \mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {3 + h} \right) - f\left( 3 \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to {0^ + }} \frac{{\left( { - {{\left( {3 + h} \right)}^2} + 9} \right) - \left( 0 \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to {0^ + }} \frac{{h\left( {6 + h} \right)}}{h}\\ &= 6\end{align}\)

Comparing results.

From above calculation, we have:

\(f_ - ^{'}\left( 3 \right) \ne f_ + ^{'}\left( 3 \right)\)

Thus, the function \(f'\left( 3 \right)\) does not exist. Similarly, \(f'\left( { - 3} \right)\) does not exist

Hence, the function \(f\) is not differential at \(x = \pm 3\).

(b) Step 5: Plotting Graph.

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

\(f\left( x \right) \Rightarrow \)

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

Hence, these are the required graphs.