1. Find the intervals of increase or decrease.
2. Find the local maximum and minimum values.
3. Find the intervals of concavity and the inflection points.
4. Use the information from parts (a)-(c) to sketch the graph. You may want to check your work with a graphing calculator or computer.

48. \(g\left( x \right) = 200 + 8{x^3} + {x^4}\)

Theincreasing and decreasing testas shown below:

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

a)

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

\(\begin{array}{c}g'\left( x \right) = 0 + 24{x^2} + 4{x^3}\\ = 4{x^2}\left( {6 + x} \right)\end{array}\)

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

\(\begin{array}{c}4{x^2}\left( {6 + x} \right) = 0\\x = 0, - 6\end{array}\)

The critical number is \(x = 0\) and \(x = - 6\).

Thus, the function \(g\) is increasing on the interval \(\left( { - 6,\infty } \right)\). The function \(g\) is decreasing on the interval \(\left( { - \infty , - 6} \right)\) and horizontal tangent at \(x = 0\).

b)

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

1. The function \(f\) has a local maximumat \(c\) when \(f'\) changes from positive to negative at \(c\).
2. The function \(f\) has a local 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 \(g\) from decreasing to increasing at \(x = - 6\).

Thus, the local minimum value is \(g\left( { - 6} \right) = - 232\) and the local maximum value does not exist.

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:

\(\begin{array}{c}g''\left( x \right) = 48x + 12x\\ = 12x\left( {4 + x} \right)\end{array}\)

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

\(\begin{array}{c}12x\left( {4 + x} \right) = 0\\x = 0, - 4\end{array}\)

The critical point of \(g\) is \(x = 0, - 4\). \(g''\left( x \right) > 0\) on the interval \(\left( { - \infty , - 4} \right)\) and \(\left( {0,\infty } \right)\). \(g''\left( x \right) < 0\) on the interval \(\left( { - 4,0} \right)\).

Thus, the function \(g\) is concave upward on the interval \(\left( { - \infty , - 4} \right)\) and \(\left( {0,\infty } \right)\). The function \(g\) is concave downward on the interval \(\left( { - 4,0} \right)\). The points\(\left( { - 4, - 56} \right)\) and \(\left( {0,200} \right)\) is an inflection point.

d)

The points\(\left( { - 4, - 56} \right)\) and \(\left( {0,200} \right)\) is an inflection point. The local minimum value is \(g\left( { - 6} \right) = - 232\). The function \(g\) is increasing on the interval \(\left( { - 6,\infty } \right)\) . The function \(g\) is decreasing on the interval \(\left( { - \infty , - 6} \right)\) and horizontal tangent at \(x = 0\).

Use the above conditions to sketch the graph of \(g\) as shown below:

Thus, the graph of the function \(g\) 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 “STAT PLOT” and enter the equation\(200 + 8{X^3} + {X^4}\)in the\({Y_1}\)tab.
2. Enter the “GRAPH” 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 \(g\) is confirmed.