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 ) 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.
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the function ( f ) is concave upward 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 concave upward.
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the function ( f ) is concave downward on the subinterval(s)
b. the function ( f ) is never concave downward.

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 \(y=-x^{4}+30x^{2}+216\)
\(f^\prime(x)=-4x^{3}+60x=-4x(x^{2}-15)=-4x(x-\sqrt{15})(x + \sqrt{15})\)
Set \(f^\prime(x)=0\), we get \(x = 0,x=\sqrt{15},x=-\sqrt{15}\)
Use the first - derivative test:

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

Since the function changes from decreasing (\(-\sqrt{15}

Step3: Find the second - derivative

Differentiate \(f^\prime(x)=-4x^{3}+60x\)
\(f^{\prime\prime}(x)=-12x^{2}+60=-12(x^{2}-5)=-12(x-\sqrt{5})(x + \sqrt{5})\)
Set \(f^{\prime\prime}(x)=0\), we get \(x=\sqrt{5},x = -\sqrt{5}\)

• For \(x<-\sqrt{5}\), let \(x=-3\), \(f^{\prime\prime}(-3)=-12\times((-3)^{2}-5)=-12\times4<0\)
• For \(-\sqrt{5}
• For \(x>\sqrt{5}\), let \(x = 3\), \(f^{\prime\prime}(3)=-12\times(9 - 5)=-48<0\)

Answer:

For the local - minimum question: A. The function \(f\) has a local minimum at \(x = 0\)
For the concave - upward question: A. The function \(f\) is concave upward on the subinterval(s) \((-\sqrt{5},\sqrt{5})\)
For the concave - downward question: A. The function \(f\) is concave downward on the subinterval(s) \((-\infty,-\sqrt{5})\cup(\sqrt{5},\infty)\)