How do you find all local maximum and minimum points using the second derivative test given #y=2+3xx^3#?
maximum at (1,4) and minimum at (1,0)
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To find all local maximum and minimum points using the second derivative test for the function ( y = 2 + 3x  x^3 ):

Find the first derivative ( y' ) by differentiating the function with respect to ( x ). [ y' = 3  3x^2 ]

Find the critical points by setting the first derivative equal to zero and solving for ( x ). [ 3  3x^2 = 0 ] [ x^2 = 1 ] [ x = \pm 1 ]

Find the second derivative ( y'' ) by differentiating ( y' ) with respect to ( x ). [ y'' = 6x ]

Evaluate the second derivative at the critical points found in step 2. [ y''(1) = 6 ] [ y''(1) = 6 ]

Analyze the signs of the second derivative at the critical points:
 If ( y''(c) > 0 ), the function has a local minimum at ( c ).
 If ( y''(c) < 0 ), the function has a local maximum at ( c ).
 If ( y''(c) = 0 ), the test is inconclusive.

Determine the nature of the critical points:
 At ( x = 1 ), ( y''(1) = 6 > 0 ), so there is a local minimum at ( x = 1 ).
 At ( x = 1 ), ( y''(1) = 6 < 0 ), so there is a local maximum at ( x = 1 ).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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