How do you find the critical points for #y=x+2x^-1#?

Question

Emily Ammerman · Accepted Answer

#f(x) y=x+2x^-1=x+2/x# #f'(x)=y'=1-2/x^2=(x^2-2)/x^2# Critical points are points in the domain at which #f'(x)=0# or #f'(x)# does not exist #f'(x)=0# when #x^2-2=0#, so #x= +- sqrt2#   which are both in the domain of #f#. #f'(x)# does not exist at #x=0# which is not in the domain of #f#, hence is not a critical point. The critical points for #f(x) y=x+2x^-1=x+2/x# are #+- sqrt2#

Evelyn Agard · Answer

undefined To find the critical points of $ y = x + 2x^{-1} $, follow these steps:

1. Find the first derivative of the function: $ y' = 1 - 2x^{-2} $.
2. Set the first derivative equal to zero and solve for $ x $: $ 1 - 2x^{-2} = 0 $.
3. Solve for $ x $ in the equation: $ 2x^{-2} = 1 $.
4. Take the reciprocal of both sides: $ x^2 = 2 $.
5. Solve for $ x $: $ x = \pm \sqrt{2} $.

Therefore, the critical points of the function are $ x = \sqrt{2} $ and $ x = -\sqrt{2} $.

William Abdalla · Answer

undefined To find the critical points for the function $ y = x + 2x^{-1} $, first find its derivative, set it equal to zero, and solve for $ x $. Then, check these values to ensure they are valid critical points.

1. Find the derivative of the function:
$$ y' = 1 - 2x^{-2} $$

2. Set the derivative equal to zero:
$$ 1 - 2x^{-2} = 0 $$

3. Solve for $ x $:
$$ 1 = 2x^{-2} $$
$$ x^{-2} = \frac{1}{2} $$
$$ x^2 = 2 $$
$$ x = \pm \sqrt{2} $$

4. Check these values to ensure they are valid critical points by plugging them into the original function or its derivative.

Therefore, the critical points are $ x = \sqrt{2} $ and $ x = -\sqrt{2} $.