# How do you find the local max and min for #f(x) = 1 - sqrt(x)#?

Local maximum is

There are no other critical numbers, so there are no other local extrema.

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To find the local maximum and minimum points for ( f(x) = 1 - \sqrt{x} ):

- Find the first derivative of the function, ( f'(x) ).
- Set ( f'(x) = 0 ) and solve for ( x ).
- Determine the critical points.
- Use the second derivative test or the first derivative test to determine if each critical point corresponds to a local maximum, minimum, or neither.

Let's start with step 1:

[ f(x) = 1 - \sqrt{x} ]

[ f'(x) = \frac{d}{dx}(1 - \sqrt{x}) = 0 - \frac{1}{2\sqrt{x}} = -\frac{1}{2\sqrt{x}} ]

Now, for step 2:

[ f'(x) = 0 ]

[ -\frac{1}{2\sqrt{x}} = 0 ]

[ \frac{1}{2\sqrt{x}} = 0 ]

[ \text{No solution.} ]

Since there are no solutions for ( f'(x) = 0 ), there are no critical points.

Therefore, there are no local maximum or minimum points for the function ( f(x) = 1 - \sqrt{x} ).

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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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