Question

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

55. \(f\left( x \right) = 3{x^4} - 4{x^3} - 12{x^2} + 1,{\rm{ }}\left( { - 2,3} \right)\)

Short answer

The absolute maximum value and absolute minimum value of the function \(f\) are \(f\left( { - 2} \right) = 33\) and \(f\left( 2 \right) = - 31\) 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 as the 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:

\(\begin{aligned}{c}f'\left( x \right) &= 3\left( 4 \right){x^3} - 4\left( 3 \right){x^2} - 12\left( 2 \right)x + 0\\ &= 12{x^3} - 12{x^2} - 24x\\ &= 12x\left( {{x^2} - x - 2} \right)\end{aligned}\)

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

\(\begin{aligned}{c}f'\left( x \right) &= 0\\12x\left( {{x^2} - x - 2} \right) &= 0\\x\left( {{x^2} - x - 2} \right) &= 0\\x\left( {x - 2} \right)\left( {x + 1} \right) &= 0\\x &= - 1,0,2\end{aligned}\)

The critical numbers of \(f\) are \(x = - 1\), \(x = 0\) and \(x = 2\).

Determine the absolute maximum and absolute minimum value

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

\(\begin{aligned}{c}f\left( { - 1} \right) &= 3{\left( { - 1} \right)^4} - 4{\left( { - 1} \right)^3} - 12{\left( { - 1} \right)^2} + 1\\ &= 3 + 4 - 12 + 1\\ &= - 4\\f\left( 0 \right) &= 3{\left( 0 \right)^4} - 4{\left( 0 \right)^3} - 12{\left( 0 \right)^2} + 1\\ &= 0 - 0 - 0 + 1\\ &= 1\\f\left( 2 \right) &= 3{\left( 2 \right)^4} - 4{\left( 2 \right)^3} - 12{\left( 2 \right)^2} + 1\\ &= 48 - 32 - 48 + 1\\ &= - 31\end{aligned}\)

Determine the values of \(f\) at the endpoints of the interval \(\left( { - 2,3} \right)\)as shown below:

\(\begin{aligned}{c}f\left( { - 2} \right) &= 3{\left( { - 2} \right)^4} - 4{\left( { - 2} \right)^3} - 12{\left( { - 2} \right)^2} + 1\\ &= 48 + 32 - 48 + 1\\ &= 33\\f\left( 3 \right) &= 3{\left( 3 \right)^4} - 4{\left( 3 \right)^3} - 12{\left( 3 \right)^2} + 1\\ &= 243 - 108 - 108 + 1\\ &= 28\end{aligned}\)

It is observed from these values of \(f\) that the absolute maximum value is\(f\left( { - 2} \right) = 33\) and the absolute minimum value is \(f\left( 2 \right) = - 31\).

Thus, the absolute maximum value and absolute minimum value of the function \(f\) are \(f\left( { - 2} \right) = 33\) and \(f\left( 2 \right) = - 31\) respectively.