# How do you find the first three iterate of the function #f(x)=2x^2-5# for the given initial value #x_0=-1#?

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To find the first three iterates of the function ( f(x) = 2x^2 - 5 ) for the initial value ( x_0 = -1 ), you can use the iterative formula ( x_{n+1} = f(x_n) ).

- Calculate ( f(x_0) ) which gives you the first iterate ( x_1 ).
- Then, plug ( x_1 ) into the function to find ( f(x_1) ), giving you the second iterate ( x_2 ).
- Repeat this process one more time to find the third iterate ( x_3 ).

Starting with ( x_0 = -1 ):

- ( f(-1) = 2(-1)^2 - 5 = 2 - 5 = -3 )
- ( f(-3) = 2(-3)^2 - 5 = 18 - 5 = 13 )
- ( f(13) = 2(13)^2 - 5 = 338 - 5 = 333 )

So, the first three iterates are: ( x_1 = -3 ), ( x_2 = 13 ), and ( x_3 = 333 ).

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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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