The function given by $f(x)=\left(x^{2}-2\right)^{3}$ 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
The positive critical number rounded to three decimal places is \( \boxed{1.414} \)
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 \( f'(x) = 6x(x^2 - 2)^2 \)
Step 4 ：We set the derivative equal to zero to find the critical points: \( 6x(x^2 - 2)^2 = 0 \)
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 \( \boxed{1.414} \)
