Question

Let \(f\left( x \right) = {\left( {x - 1} \right)^2}\), \(g\left( x \right) = {e^{ - 2x}}\)and \(h\left( x \right) = 1 + \ln \left( {1 - 2x} \right)\)

(a) Find the linearization of \(f\), \(g\), and \(h\)at \(a = 0\). What do you notice? How do you explain what happened?

(b) Graph \(f\), \(g\), and \(h\) and their linear approximations. For which function is the linear approximation best? For which is it worst? Explain.

Short answer

1. The linearization is \({L_f} = {L_h} = {L_g} = 1 - 2x\). This is because the function \(f,\,g,\,h\) have same definition and derivative at \(x = 0\).
2. The graph is shown below:

The linear approximation is best suited for function \(f\) as the function appears to be linear for a larger domain as compared to \(g\)and \(h\). The approximation is not suited for \(h\) as \(h\) deviates from \(L\) at larger extent compared to \(g\)and \(f\).

Step-by-step solution

Find linearization of \(f\), \(g\) and \(h\)

The derivative of function \(f\left( x \right) = {\left( {x - 1} \right)^2}\) is \(f'\left( x \right) = 2\left( {x - 1} \right)\) such that \(f\left( 0 \right) = 1\) and \(f'\left( 0 \right) = - 2\). Thus the linearization of function \({L_f}\left( x \right) = f\left( 0 \right) + f'\left( 0 \right)\left( {x - 0} \right)\). Plugging in all the values, we get:

^.\(\begin{aligned}{L_f}\left( x \right) &= 1 + \left( { - 2} \right)\left( {x - 0} \right)\\ &= 1 - 2x\end{aligned}\)

The derivative of function \(g\left( x \right) = {e^{ - 2x}}\) is \(g'\left( x \right) = - 2{e^{ - 2x}}\)such that \(g\left( 0 \right) = 1\) and \(g'\left( 0 \right) = - 2\). Thus the linearization of function \({L_g}\left( x \right) = g\left( 0 \right) + g'\left( 0 \right)\left( {x - 0} \right)\). Plugging in all the values, we get:

^.\(\begin{aligned}{g_f}\left( x \right) &= 1 + \left( { - 2} \right)\left( {x - 0} \right)\\ &= 1 - 2x\end{aligned}\)

The derivative of function \(h\left( x \right) = 1 + \ln \left( {1 - 2x} \right)\) is \(h'\left( x \right) = \frac{{ - 2}}{{1 - 2x}}\) such that \(h\left( 0 \right) = 1\)and \(h'\left( 0 \right) = - 2\). Thus the linearization of function \({L_h}\left( x \right) = h\left( 0 \right) + h'\left( 0 \right)\left( {x - 0} \right)\). Plugging in all the values, we get:

^.\(\begin{aligned}{L_h}\left( x \right) &= 1 + \left( { - 2} \right)\left( {x - 0} \right)\\ &= 1 - 2x\end{aligned}\)

It can be noticed \({L_f} = {L_h} = {L_g}\). This is because the function \(f,\,g,\,h\) have same definition and derivative at \(x = 0\).

Graph the functions

Draw the graph of the functions\(L\left( x \right) = 1 - 2x\)\(f\left( x \right) = {\left( {x - 1} \right)^2}\), \(g\left( x \right) = {e^{ - 2x}}\), \(h\left( x \right) = 1 + \ln \left( {1 - 2x} \right)\)by using the graphing calculator as shown below:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\({\left( {X - 1} \right)^2}\)in the\({Y_1}\)tab,\({e^{ - 2X}}\)in the\({Y_2}\)tab,\(1 + \ln \left( {1 - 2X} \right)\)in the\({Y_3}\)tab,\(1 - 2X\)in the\({Y_4}\)tab.
2. Enter the “GRAPH” button in the graphing calculator.

Visualization of graph of the functions is shown below:

Draw the conclusion

The linear approximation is best suited for function \(f\) as the function appears to be linear for a larger domain as compared to \(g\)and \(h\). The approximation is not suited for \(h\) as \(h\) deviates from \(L\) at larger extent compared to \(g\) and \(f\).