Question

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( y = f(x) ).
( f(x)=left(x^{2}+6
ight)left(36 - x^{2}
ight) )
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 = -sqrt{15},sqrt{15} ).
(type an exact answer, using radicals as needed. use a comma to separate answers as needed.)
b. the function ( f ) has no local maximum.
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the function ( f ) has a local minimum 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 minimum.

Explanation:

Step1: Expand the function

First, expand \(f(x)=(x^{2}+6)(36 - x^{2})\). Using the FOIL method:

$$
LATEXBLOCK0
$$

Step2: Find the first - derivative

Differentiate \(y = f(x)=-x^{4}+30x^{2}+216\) with respect to \(x\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), we have \(y^\prime=f^\prime(x)=-4x^{3}+60x=-4x(x^{2}-15)=-4x(x-\sqrt{15})(x + \sqrt{15})\)

Step3: Determine the critical points

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

Step4: Use the first - derivative test

• Interval \((-\infty,-\sqrt{15})\): Let \(x=-4\). Then \(f^\prime(-4)=-4\times(-4)\times((-4)^{2}-15)=16\times1 = 16>0\), so \(f(x)\) is increasing on \((-\infty,-\sqrt{15})\)
• Interval \((-\sqrt{15},0)\): Let \(x=-1\). Then \(f^\prime(-1)=-4\times(-1)\times((-1)^{2}-15)=4\times(-14)=-56<0\), so \(f(x)\) is decreasing on \((-\sqrt{15},0)\)
• Interval \((0,\sqrt{15})\): Let \(x = 1\). Then \(f^\prime(1)=-4\times1\times(1^{2}-15)=-4\times(-14)=56>0\), so \(f(x)\) is increasing on \((0,\sqrt{15})\)
• Interval \((\sqrt{15},\infty)\): Let \(x = 4\). Then \(f^\prime(4)=-4\times4\times(4^{2}-15)=-16\times1=-16<0\), so \(f(x)\) is decreasing on \((\sqrt{15},\infty)\)

Since the function changes from decreasing to increasing at \(x = 0\), by the first - derivative test, the function \(f(x)\) has a local minimum at \(x = 0\)

Answer:

For the local minimum:
A. The function \(f\) has a local minimum at \(x = 0\)