The function given by $f (x) = {(x^{2} - 2)}^{3}$ has exactly one positive critical number. What is that positive critical number?

Note: We are asking for the $x$ -coordinate.
Enter your answer as a decimal. Round to three decimal places (as needed).

Answer

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Answer

The positive critical number rounded to three decimal places is $1.414$

Steps

Step 1 ：The function is given by $f (x) = (x^{2} - 2)^{3}$

Step 2 ：To find the critical number, we need to take the derivative of the function and set it equal to zero.

Step 3 ：The derivative of the function is $f^{'} (x) = 6 x (x^{2} - 2)^{2}$

Step 4 ：We set the derivative equal to zero to find the critical points: $6 x (x^{2} - 2)^{2} = 0$

Step 5 ：The critical points are $x = 0$ , $x = - \sqrt{2}$ , and $x = \sqrt{2}$

Step 6 ：We are looking for the positive critical number, which is $x = \sqrt{2}$

Step 7 ：The positive critical number rounded to three decimal places is $1.414$