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 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 = 0 ).
(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
ob. the function f is never concave upward.

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 \((x^{n})^\prime=nx^{n - 1}\), we have \(f^\prime(x)=-4x^{3}+60x=-4x(x^{2}-15)=-4x(x-\sqrt{15})(x + \sqrt{15})\)
Set \(f^\prime(x)=0\), then \(x = 0,x=\sqrt{15},x=-\sqrt{15}\)

Step3: Use the first - derivative test

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

Since \(f^\prime(x)\) changes sign from positive to negative at \(x=-\sqrt{15}\) and \(x=\sqrt{15}\), the function \(f(x)\) has local maxima at \(x =-\sqrt{15}\) and \(x=\sqrt{15}\)

Since \(f^\prime(x)\) changes sign from negative to positive at \(x = 0\), the function \(f(x)\) has a local minimum at \(x = 0\)

Step4: Find the second - derivative

\(f^{\prime\prime}(x)=-12x^{2}+60=-12(x^{2}-5)=-12(x-\sqrt{5})(x + \sqrt{5})\)
Set \(f^{\prime\prime}(x)=0\), then \(x=\pm\sqrt{5}\)

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

The function \(f(x)\) is concave upward on the interval \((-\sqrt{5},\sqrt{5})\)

Answer:

For the local maximum: A. The function \(f\) has a local maximum at \(x=-\sqrt{15},\sqrt{15}\)
For the local minimum: A. The function \(f\) has a local minimum at \(x = 0\)
For the concavity: A. The function \(f\) is concave upward on the sub - interval\((-\sqrt{5},\sqrt{5})\)