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 ( x )-intercept(s)? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the ( x )-intercept(s) is/are ( x=-sqrt{6}, 0, sqrt{6} ).
(type an exact answer, using radicals as needed. use a comma to separate answers as needed.)
b. there is no ( x )-intercept.
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 comma to sepa
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=square ).
(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.

Explanation:

Step1: Find the derivative of \(f(x)\)

Given \(f(x)=2x^{4}-12x^{2}\), then \(f^{\prime}(x)=8x^{3}-24x = 8x(x^{2}-3)=8x(x - \sqrt{3})(x+\sqrt{3})\)

Step2: Find the critical points

Set \(f^{\prime}(x)=0\), so \(8x(x - \sqrt{3})(x+\sqrt{3})=0\). The critical points are \(x = 0,x=\sqrt{3},x =-\sqrt{3}\)

Step3: Use the second - derivative test

First, find the second - derivative \(f^{\prime\prime}(x)=24x^{2}-24\)

• When \(x = 0\), \(f^{\prime\prime}(0)=24\times0^{2}-24=- 24<0\), so \(x = 0\) is a local maximum.
• When \(x=\sqrt{3}\), \(f^{\prime\prime}(\sqrt{3})=24\times(\sqrt{3})^{2}-24=24\times3 - 24=48>0\), so \(x=\sqrt{3}\) is a local minimum.
• When \(x=-\sqrt{3}\), \(f^{\prime\prime}(-\sqrt{3})=24\times(-\sqrt{3})^{2}-24=24\times3 - 24=48>0\), so \(x=-\sqrt{3}\) is a local minimum.

Answer:

The local minimum/a is/are at \(x=-\sqrt{3},\sqrt{3}\)