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 ( x )-intercept(s) of ( f ) is/are ( x = -6,6 ).
(type an exact answer, using radicals as needed. use a comma to separate answers as needed.)
b. the function ( f ) has no ( x )-intercepts.
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the ( y )-intercept of ( f ) is ( y = 216 ).
(type an exact answer, using radicals as needed.)
b. the function ( f ) has no ( y )-intercept.
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)
(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.

Explanation:

Step1: Find the x - intercepts

Set \(y = f(x)=0\), so \((x^{2}+6)(36 - x^{2})=0\).
Since \(x^{2}+6>0\) for all real \(x\) (because \(x^{2}\geq0\), then \(x^{2}+6\geq6\)), we solve \(36 - x^{2}=0\).
Using the difference of squares formula \(a^{2}-b^{2}=(a + b)(a - b)\), where \(a = 6\) and \(b=x\), we have \((6 + x)(6 - x)=0\).
Set each factor equal to zero: \(6+x=0\) gives \(x=-6\), and \(6 - x=0\) gives \(x = 6\).

Step2: Find the y - intercepts

Set \(x = 0\) in \(y=f(x)\). Then \(y=(0^{2}+6)(36-0^{2})=6\times36 = 216\).

Step3: Analyze the function's increasing/decreasing behavior

First, expand \(f(x)=(x^{2}+6)(36 - x^{2})=36x^{2}-x^{4}+216 - 6x^{2}=-x^{4}+30x^{2}+216\).
Take the derivative \(f^\prime(x)=-4x^{3}+60x=-4x(x^{2}-15)=-4x(x-\sqrt{15})(x + \sqrt{15})\).
Set \(f^\prime(x)=0\), we get \(x = 0,x=\sqrt{15},x=-\sqrt{15}\).
Use a sign - chart for \(f^\prime(x)\):

• When \(x<-\sqrt{15}\), let \(x=-4\), then \(f^\prime(-4)=-4\times(-4)\times((-4)^{2}-15)=16\times1>0\).
• When \(-\sqrt{15}
• When \(0
• When \(x>\sqrt{15}\), let \(x = 4\), then \(f^\prime(4)=-4\times4\times(4^{2}-15)=-16\times1<0\).

So the function \(f(x)\) is increasing on the intervals \((-\infty,-\sqrt{15})\) and \((0,\sqrt{15})\).

Answer:

For the x - intercepts: A. The x - intercept(s) of \(f\) is/are \(x=-6,6\).
For the y - intercepts: A. The y - intercept of \(f\) is \(y = 216\).
For the increasing intervals: A. The function \(f\) is increasing on the subinterval(s) \((-\infty,-\sqrt{15}),(0,\sqrt{15})\).