Question

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

58. \(f\left( x \right) = \frac{x}{{{x^2} - x + 1}},{\rm{ }}\left( {0,3} \right)\)

Short answer

The absolute maximum value and absolute minimum value of the function \(f\) are \(f\left( 1 \right) = 1\) and \(f\left( 0 \right) = 0\) respectively.

Step-by-step solution

Definition of absolute maximum and absolute minimum value

Suppose that \(c\) is the number in the domain \(D\) of a function \(f\). Then

• \(f\left( c \right)\)is known asthe absolute maximum valueof \(f\) on the domain \(D\) when \(f\left( c \right) \ge f\left( x \right)\) for all \(x\) in \(D\).
• \(f\left( c \right)\)is known asthe absolute minimum valueof \(f\) on the domain \(D\) when \(f\left( c \right) \le f\left( x \right)\) for all \(x\) in \(D\).

Determine the critical number of \(f\)

TheClosed Interval Method: To calculate the absolute maximum and absolute minimum valueof a continuous function \(f\) with the closed interval \(\left( {a,b} \right)\), use the following steps.

1. Determine the values of \(f\) at the critical number in \(\left( {a,b} \right)\).
2. Determine the values of \(f\) at the endpoints of the closed interval \(\left( {a,b} \right)\).
3. The absolute maximum value is the greatest of the values in steps 1 and 2, while the absolute minimum value is the lowest of these values.

Obtain the derivative of \(f\) as shown below:

\(\begin{aligned}{c}f'\left( x \right) &= \frac{{\left( {{x^2} - x + 1} \right)\left( 1 \right) - x\left( {2x - 1} \right)}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\\ &= \frac{{{x^2} - x + 1 - 2{x^2} + x}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\\ &= \frac{{1 - {x^2}}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\\ &= \frac{{\left( {1 + x} \right)\left( {1 - x} \right)}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\end{aligned}\)

When \(f'\left( x \right) = 0\) then the critical number of \(f\) as shown below:

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

However, \(x = - 1\) does not exist in the interval \(\left( {0,3} \right)\). Therefore, the critical numbers of \(f\) is \(x = 1\).

Determine the absolute maximum and absolute minimum value

Determine the values of \(f\) at the critical numbers as shown below:

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

Determine the values of \(f\) at the endpoints of the interval \(\left( {0,3} \right)\) as shown below:

\(\begin{aligned}{c}f\left( 0 \right) &= \frac{0}{{{{\left( 0 \right)}^2} - 0 + 1}}\\ &= 0\\f\left( 3 \right) &= \frac{3}{{{{\left( 3 \right)}^2} - 3 + 1}}\\ &= \frac{3}{{9 - 3 + 1}}\\ &= \frac{3}{7}\end{aligned}\)

It is observed from these values of \(f\) that the absolute maximum value is\(f\left( 1 \right) = 1\) and the absolute minimum value is \(f\left( 0 \right) = 0\).

Thus, the absolute maximum value and absolute minimum value of the function \(f\) are \(f\left( 1 \right) = 1\) and \(f\left( 0 \right) = 0\) respectively.