Question

Use the graphs of \(f,f',f''\) to estimate the \(x\)-coordinates of the maximum and minimum points and inflection points of \(f\).

\(f\left( x \right) = {e^{ - 0.1x}}\ln \left( {{x^2} - 1} \right)\)

Short answer

Thus, the inflection point is at \(x \approx - 4.31,11.74\), local maxima at \(x = 5.87\) and no local minima..

Step-by-step solution

Graph of the given function.

The graph for the function \(f\left( x \right) = {e^{ - 0.1x}}\ln \left( {{x^2} - 1} \right)\) is shown below:

Differentiation of the given function.

Use a calculator, the derivative of given function is as:

\(f'\left( x \right) = {e^{ - 0.1x}}\left( {\frac{{2x}}{{{x^2} - 1}} - 0.1\ln \left( {{x^2} - 1} \right)} \right)\)

From the graph, the local extrema are at \(x = 5.87\), because the sign of \(f'\left( x \right)\) changed.

From the graph of \(f\left( x \right)\),this local extrema is local maxima.

Second derivative of the given function.

Use a calculator, the second derivative of given function is as

\(f''\left( x \right) = \frac{{{e^{ - 0.1x}}\left( { - 0.4{x^3} - 2{x^2} + \left( {0.01{x^4} - 0.02{x^2} + 0.01} \right)\ln \left( {{x^2} - 1} \right) + 0.4x - 2} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}}\)

Inflection occurs when \(f''\left( x \right)\) changes its sign.

From the graph below, the inflection occurs at \(x \approx - 4.31,11.74\).

Graph of \(f\left( x \right)\) with labelled of local maxima and inflection points.

Hence, the inflection point is at \(x \approx - 4.31,11.74\), local maxima at \(x = 5.87\) and no local minima.