How do use the first derivative test to determine the local extrema #y=x(sqrt(8-x^2))#?

Answer 1

# x = +- 2, y = +-4 #

A stationary point can be obtained from # f'(x) = 0 #
# y = xsqrt(8-x^2) # # y' = sqrt(8-x^2) - x^2/sqrt(8-x^2) = (-2(x^2-4))/sqrt(8-x^2) #
# (-2(x^2-4))/sqrt(8-x^2) = 0 # # x^2 = 4 # # x = +- 2, y = +-4 #
You can then proceed to find whether these points are local maxima or minima by taking the second derivative and substituting the values of the stationary points, #alpha#: #f''(alpha) > 0 rArr# minima #f''(alpha) < 0 rArr# maxima
# y'' = -(2x(12-x^2))/(8-x^2)^(3/2) # # y''_(x=2) = -4 < 0 rArr # maxima # y''_(x=-2) = 4 > 0 rArr # minima

graph{xsqrt(8-x^2) [-5, 5, -6, 6]}

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Answer 2

To use the first derivative test to determine the local extrema of ( y = x\sqrt{8 - x^2} ):

  1. Find the first derivative of the function ( y ) with respect to ( x ).
  2. Set the derivative equal to zero and solve for ( x ).
  3. Determine the critical points.
  4. Use the first derivative test to analyze the behavior of the function around each critical point.
    • If the derivative changes sign from positive to negative at a critical point, it indicates a local maximum.
    • If the derivative changes sign from negative to positive at a critical point, it indicates a local minimum.
  5. Identify any local extrema based on the results of the first derivative test.

Let's proceed with these steps:

  1. Find the first derivative of ( y ) with respect to ( x ): [ y = x\sqrt{8 - x^2} ] [ y' = \sqrt{8 - x^2} + x\left(\frac{1}{2\sqrt{8 - x^2}}\right)(-2x) ] [ y' = \sqrt{8 - x^2} - \frac{x^2}{\sqrt{8 - x^2}} ]

  2. 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 ]

  3. Determine the critical points: ( x = 2 ) and ( x = -2 ).

  4. Use the first derivative test:

    • For ( x = 2 ): Evaluate ( y' ) for a value slightly less than 2 and a value slightly more than 2 to determine the sign change.
    • For ( x = -2 ): Evaluate ( y' ) for a value slightly less than -2 and a value slightly more than -2 to determine the sign change.
  5. Identify any local extrema based on the results of the first derivative test.

Alternatively, you can directly analyze the behavior of the function in the interval ((-∞, -2)), ((-2, 2)), and ((2, ∞)).

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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.

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.

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.

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.

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