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Theincreasing and decreasing test as shown below:

1. The function \(f\) is increasingon the interval when \(f'\left( x \right) > 0\) on an interval.
2. The function \(f\) decreases on the interval when \(f'\left( x \right) < 0\) on an interval.

a)

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

\(\begin{array}{c}f'\left( x \right) = 36 + 6x - 6{x^2}\\ = 6\left( {{x^2} - x - 6} \right)\\ = - 6\left( {x + 2} \right)\left( {x - 3} \right)\end{array}\)

Take \(f'\left( x \right) = 0\) to obtain the critical numbers as shown below:

\(\begin{array}{c} - 6\left( {x + 2} \right)\left( {x - 3} \right) = 0\\\left( {x + 2} \right)\left( {x - 3} \right) = 0\\x = - 2,3\end{array}\)

The critical number is \(x = - 2\) and \(x = 3\).

Thus, the function \(f\) is increasing on the interval \(\left( { - 2,3} \right)\). The function \(f\) is decreasing on the interval \(\left( { - \infty , - 2} \right)\) and \(\left( {3,\infty } \right)\).

b)

Thefirst derivative test: Let \(c\) be the critical number of a continuous function \(f\).

1. The function \(f\) has a local maximum at \(c\) when \(f'\) changes from positive to negative at \(c\).
2. The function \(f\) has alocal minimumat \(c\) when \(f'\) changes from negative to positive at \(c\).
3. When \(f'\) is positive to the left and right of \(c\), or negative to the left and right of \(c\), then \(f\) contain no local maximum or minimum at \(c\).

There are changes in \(f\) from increasing to decreasing at \(x = 3\) and from decreasing to increasing at \(x = - 2\).

Thus, the local maximum value is \(f\left( 3 \right) = 81\) , and the local minimum value is \(f\left( { - 2} \right) = - 44\).

c)

Theconcavity test is shown below:

1. When \(f''\left( x \right) > 0\) on an interval, \(I\)then the graph of \(f\) is said to be concave upwardon \(I\).
2. When \(f''\left( x \right) < 0\) on an interval, \(I\)then the graph of \(f\) is said to be concave downwardon \(I\).

Obtain the second derivative of the function as shown below:

\(\begin{array}{c}f''\left( x \right) = 6 - 12x\\ = 6\left( {1 - 2x} \right)\end{array}\)

Take \(f''\left( x \right) = 0\) to obtain the critical number as shown below:

\(\begin{array}{c}6\left( {1 - 2x} \right) = 0\\1 - 2x = 0\\2x = 1\\x = \frac{1}{2}\end{array}\)

The critical point of \(f\) is \(x = \frac{1}{2}\). For \(x < \frac{1}{2}\), \(f''\left( x \right) > 0\) on the interval \(\left( { - \infty ,\frac{1}{2}} \right)\) and for \(x > \frac{1}{2}\), \(f''\left( x \right) < 0\) on the interval \(\left( {\frac{1}{2},\infty } \right)\).

Thus, the function \(f\) is concave upward on \(\left( { - \infty ,\frac{1}{2}} \right)\) and \(f\) is concave downward on \(\left( {\frac{1}{2},\infty } \right)\). The point \(\left( {\frac{1}{2},\frac{{37}}{2}} \right)\) is an inflection point.

d)

The function \(f\) is increasing on the interval \(\left( { - 2,3} \right)\). The function \(f\) is decreasing on the interval \(\left( { - \infty , - 2} \right)\) and \(\left( {3,\infty } \right)\). The point \(\left( {\frac{1}{2},\frac{{37}}{2}} \right)\) is an inflection point. The local maximum value is \(f\left( 3 \right) = 81\) and the local minimum value is \(f\left( { - 2} \right) = - 44\).

Use the above condition to sketch the graph of \(f\) as shown below:

Thus, the graph of the function \(f\) is obtained.

The procedure to draw the graph of the equation by using the graphing calculator is as follows:

1. Open the graphing calculator. Select the 鈥淪TAT PLOT鈥� and enter the equation\(Y = 36X + 3{X^2} - 2{X^3}\)in the\({Y_1}\)tab.
2. Enter the 鈥淕RAPH鈥� button in the graphing calculator.

Visualization of the graph of the function as shown below:

It is observed from the graph that the graph of the function \(f\) is confirmed.