Question

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( y = f(x) ).

( f(x)=(x^{2}+6)(36 - x^{2}) )

select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. the function ( f ) is increasing on the subinterval(s) ( (-infty,-sqrt{15}),(0,sqrt{15}) ).
(type your answer in interval notation. type an exact answer, using radicals as needed. use a comma to separate answers as needed.)

b. the function ( f ) is never increasing.

select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. the function ( f ) is decreasing on the subinterval(s) ( (-sqrt{15},0),(sqrt{15},infty) ).
(type your answer in interval notation. type an exact answer, using radicals as needed. use a comma to separate answers as needed.)

b. the function ( f ) is never decreasing.

select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. the function ( f ) has a local maximum at ( x = ).
(type an exact answer, using radicals as needed. use a comma to separate answers as needed.)

b. the function ( f ) has no local maximum.

Explanation:

Step1: Expand the function

$$\begin{align*} f(x)&=(x^{2}+6)(36 - x^{2})\\ &=36x^{2}-x^{4}+216 - 6x^{2}\\ &=-x^{4}+30x^{2}+216 \end{align*}$$

Step2: Find the first - derivative

Using the power rule \(y = ax^{n}\), \(y^\prime=anx^{n - 1}\), for \(y=-x^{4}+30x^{2}+216\), we have \(f^\prime(x)=-4x^{3}+60x=-4x(x^{2}-15)=-4x(x-\sqrt{15})(x + \sqrt{15})\)

Step3: Find the critical points

Set \(f^\prime(x)=0\), then \(-4x(x-\sqrt{15})(x + \sqrt{15})=0\). The critical points are \(x = 0\), \(x=\sqrt{15}\), and \(x=-\sqrt{15}\)

Step4: Determine the intervals of increase and decrease

• Test the interval \((-\infty,-\sqrt{15})\):

Let \(x=-4\), then \(f^\prime(-4)=-4\times(-4)\times((-4)^{2}-15)=16\times1>0\). So \(f(x)\) is increasing on \((-\infty,-\sqrt{15})\)

• Test the interval \((-\sqrt{15},0)\):

Let \(x=-1\), then \(f^\prime(-1)=-4\times(-1)\times((-1)^{2}-15)=4\times(-14)<0\). So \(f(x)\) is decreasing on \((-\sqrt{15},0)\)

• Test the interval \((0,\sqrt{15})\):

Let \(x = 1\), then \(f^\prime(1)=-4\times1\times(1^{2}-15)=-4\times(-14)>0\). So \(f(x)\) is increasing on \((0,\sqrt{15})\)

• Test the interval \((\sqrt{15},\infty)\):

Let \(x = 4\), then \(f^\prime(4)=-4\times4\times(4^{2}-15)=-16\times1<0\). So \(f(x)\) is decreasing on \((\sqrt{15},\infty)\)

Step5: Find local maxima

Since the function changes from increasing to decreasing at \(x=-\sqrt{15}\) and \(x = \sqrt{15}\)
\(f(-\sqrt{15})=-(-\sqrt{15})^{4}+30(-\sqrt{15})^{2}+216=-225 + 450+216=441\)
\(f(\sqrt{15})=-(\sqrt{15})^{4}+30(\sqrt{15})^{2}+216=-225 + 450+216=441\)

Answer:

• For the increasing intervals: \(A\). The function \(f\) is increasing on the sub - interval(s) \((-\infty,-\sqrt{15}),(0,\sqrt{15})\)
• For the decreasing intervals: \(A\). The function \(f\) is decreasing on the sub - interval(s) \((-\sqrt{15},0),(\sqrt{15},\infty)\)
• For local maxima: \(A\). The function \(f\) has a local maximum at \(x=-\sqrt{15},\sqrt{15}\)