# How do you find all local maximum and minimum points using the second derivative test given #y=4x+sqrt(1-x)#?

At local maximum and minimum points the tangent is zero. We can use this to find where these points are. We first find the first derivative and equate this to zero.

Given

Equate to zero:

Using the second derivative we know that if:

Plugging in

This is confirmed by the graph:

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To find all local maximum and minimum points using the second derivative test for ( y = 4x + \sqrt{1-x} ), follow these steps:

- Find the first derivative ( y' ) of the function.
- Find the second derivative ( y'' ) of the function.
- Set ( y' = 0 ) and solve for ( x ) to find critical points.
- Use the second derivative test to determine if the critical points are local maxima, local minima, or neither.

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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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