Question

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( y = f(x) ).
( f(x)=2 x^{4}-12 x^{2} )
what is/are the local maximum/a? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the local maximum/a is/are at ( x = 0 ).
(simplify your answer. type an exact answer using radicals as needed. use integers or fractions for any numbers in the expression. use a comm
b. there is no local maximum.
what is/are the local minimum/a? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the local minimum/a is/are at ( x = -sqrt{3}, sqrt{3} ).
(simplify your answer. type an exact answer using radicals as needed. use a comma to separate answers as needed)
b. there is no local minimum.
what are the inflection points? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the inflection points are at ( x = )
(simplify your answer. type an exact answer using radicals as needed. use a comma to separate answers as needed)
b. there are no inflection points

Explanation:

Step1: Find the first derivative

The function is \(f(x) = 2x^{4}-12x^{2}\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the first derivative \(f^\prime(x)=8x^{3}-24x=8x(x^{2}-3)=8x(x-\sqrt{3})(x + \sqrt{3})\).
Set \(f^\prime(x)=0\), we get \(x = 0,x=\sqrt{3},x=-\sqrt{3}\).
Use the first - derivative test:

• For \(x\lt-\sqrt{3}\), let \(x=-2\), then \(f^\prime(-2)=8\times(-2)\times((-2)^{2}-3)=-16\lt0\).
• For \(-\sqrt{3}\lt x\lt0\), let \(x = - 1\), then \(f^\prime(-1)=8\times(-1)\times((-1)^{2}-3)=16\gt0\).
• For \(0\lt x\lt\sqrt{3}\), let \(x = 1\), then \(f^\prime(1)=8\times1\times(1^{2}-3)=-16\lt0\).
• For \(x\gt\sqrt{3}\), let \(x = 2\), then \(f^\prime(2)=8\times2\times(2^{2}-3)=16\gt0\).

So \(x = 0\) is a local maximum.

Step2: Find the second derivative

Differentiate \(f^\prime(x)=8x^{3}-24x\) using the power rule. \(f^{\prime\prime}(x)=24x^{2}-24 = 24(x^{2}-1)=24(x - 1)(x + 1)\).
Set \(f^{\prime\prime}(x)=0\), we get \(x=-1,x = 1\).

• When \(x\lt-1\), let \(x=-2\), \(f^{\prime\prime}(-2)=24\times((-2)^{2}-1)=72\gt0\).
• When \(-1\lt x\lt1\), let \(x = 0\), \(f^{\prime\prime}(0)=24\times(0^{2}-1)=-24\lt0\).
• When \(x\gt1\), let \(x = 2\), \(f^{\prime\prime}(2)=24\times(2^{2}-1)=72\gt0\).

Answer:

The inflection points are at \(x=-1,1\). So the answer is \(A\). The inflection points are at \(x=-1,1\).