Home > Calculus > Identifying Stationary Points (Critical Points) for a Function

How do you find the critical points for #y = x*sqrt(8-x^2)#?

Answer 1

Cameron Alexander

Refer Explanation

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Aurora Alvarado

To find the critical points of ( y = x\sqrt{8 - x^2} ), we need to first find the derivative of ( y ) with respect to ( x ), set it equal to zero, and solve for ( x ).

Let's denote ( y = x\sqrt{8 - x^2} ).

Using the product rule, the derivative of ( y ) with respect to ( x ) is:

[ \frac{dy}{dx} = \sqrt{8 - x^2} + x \left( \frac{1}{2\sqrt{8 - x^2}} \cdot (-2x) \right) ]

[ \frac{dy}{dx} = \sqrt{8 - x^2} - \frac{x^2}{\sqrt{8 - x^2}} ]

To find the critical points, we set the derivative equal to zero and solve for ( x ):

[ \sqrt{8 - x^2} - \frac{x^2}{\sqrt{8 - x^2}} = 0 ]

[ \sqrt{8 - x^2} = \frac{x^2}{\sqrt{8 - x^2}} ]

[ (8 - x^2) = x^2 ]

[ 8 = 2x^2 ]

[ x^2 = 4 ]

[ x = \pm 2 ]

So, the critical points are ( x = -2 ) and ( x = 2 ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.