# How do you find all local maximum and minimum for #g(x)=6 x^3−(144 )x^2 +(1080 )x−4#?

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To find all local maximum and minimum for the function ( g(x) = 6x^3 - 144x^2 + 1080x - 4 ):

- Take the derivative of the function to find critical points.
- Set the derivative equal to zero and solve for ( x ) to find critical points.
- Once you have critical points, use the second derivative test to determine whether each critical point is a local maximum, local minimum, or neither.

Let's find the derivative of ( g(x) ): [ g'(x) = 18x^2 - 288x + 1080 ]

Now, set the derivative equal to zero and solve for ( x ): [ 18x^2 - 288x + 1080 = 0 ] [ 2x^2 - 32x + 120 = 0 ]

Divide the equation by 2: [ x^2 - 16x + 60 = 0 ]

Factor the quadratic equation: [ (x - 10)(x - 6) = 0 ]

So, the critical points are ( x = 10 ) and ( x = 6 ).

Now, find the second derivative of ( g(x) ): [ g''(x) = 36x - 288 ]

Evaluate the second derivative at each critical point: [ g''(6) = 36(6) - 288 = -72 ] [ g''(10) = 36(10) - 288 = 72 ]

Since ( g''(6) ) is negative, ( x = 6 ) is a local maximum. Since ( g''(10) ) is positive, ( x = 10 ) is a local minimum.

So, the local maximum occurs at ( x = 6 ) and the local minimum occurs at ( x = 10 ).

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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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