Question

Find the absolute maximum and absolute minimum values of \(f\) on the given interval.

65. \(f\left( x \right) = \ln \left( {{x^2} + x + 1} \right),\,\,\left( { - 1,1} \right)\)

Short answer

The absolute maximum value of the function \(f\left( x \right) = \ln \left( {{x^2} + x + 1} \right)\)is \(f\left( 1 \right) \approx 1.099\).

The absolute minimum value of the function \(f\left( x \right) = \ln \left( {{x^2} + x + 1} \right)\) is \(f\left( { - \frac{1}{2}} \right) \approx - 0.288\).

Step-by-step solution

Absolute Extremum

Suppose, a continuous function on a closed interval\(\left( {a,b} \right)\)say,\(f\left( x \right)\). Then, the function has both absolute maximum and absolute minimum. Which is given by obtained the critical numbers and then, find\(f\left( x \right)\)at the end points and the critical numbers then, compare the values. The maximum value will give the absolute maxima and the minimum value gives absolute minima.

Find the critical points by differentiating the given function

The given function is \(f\left( x \right) = \ln \left( {{x^2} + x + 1} \right)\).

So, differentiating the above function with respect to \(x\) as:

\(\begin{aligned}{c}\frac{d}{{dx}}\left( {f\left( x \right)} \right) &= \frac{d}{{dx}}\left( {\ln \left( {{x^2} + x + 1} \right)} \right)\\f'\left( x \right) &= \frac{1}{{{x^2} + x + 1}} \cdot \frac{d}{{dx}}\left( {{x^2} + x + 1} \right)\\ &= \frac{{\frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( 1 \right)}}{{{x^2} + x + 1}}\\ &= \frac{{2x + 1 + 0}}{{{x^2} + x + 1}}\\ &= \frac{{2x + 1}}{{{x^2} + x + 1}}\end{aligned}\)

Now, substitute \(f'\left( x \right) = 0\) and solve for \(x\) to get the critical points.

\(\begin{aligned}{c}\frac{{2x + 1}}{{{x^2} + x + 1}} &= 0\\2x + 1 &= 0\\x &= - \frac{1}{2}\end{aligned}\)

Find value of \(f\) at critical points in \(\left( { - 1,1} \right)\)

At \(x = - \frac{1}{2}\),

\(\begin{aligned}{c}f\left( { - \frac{1}{2}} \right) &= \ln \left( {{{\left( { - \frac{1}{2}} \right)}^2} + \left( { - \frac{1}{2}} \right) + 1} \right)\\ &= \ln \left( {\frac{3}{4}} \right)\\ &\approx - 0.288\end{aligned}\)

Find value of \(f\) at endpoints of \(\left( { - 1,1} \right)\)

At \(x = - 1\)

\(\begin{aligned}{c}f\left( { - 1} \right) &= \ln \left( {{{\left( { - 1} \right)}^2} + \left( { - 1} \right) + 1} \right)\\ &= \ln \left( 1 \right)\\ &= 0\end{aligned}\)

At \(x = 1\)

\(\begin{aligned}{c}f\left( 1 \right) &= \ln \left( {{{\left( 1 \right)}^2} + 1 + 1} \right)\\ &= \ln \left( 3 \right)\\ &\approx 1.099\end{aligned}\)

Now, find the absolute maximum and minimum value from the above steps

By the above two steps maximum and minimum value can be as when \(x = - \frac{1}{2}\)the value of \(f\left( { - \frac{1}{2}} \right) \approx - 0.288\) that is it is the minimum value and when \(x = 1\)the value of \(f\left( 1 \right) \approx 1.099\) that is it is the maximum value.

Thus, the absolute maximum and minimum values of the function \(f\left( x \right) = \ln \left( {{x^2} + x + 1} \right)\)are \(f\left( 1 \right) \approx 1.099\) and \(f\left( { - \frac{1}{2}} \right) \approx - 0.288\) respectively.