# How do you use a linear approximation or differentials to estimate #(2.001)^5#?

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To estimate ( (2.001)^5 ) using linear approximation or differentials, you can use the formula:

[ f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x ]

Where:

- ( f(x) = x^5 ) is the function you want to approximate.
- ( x = 2 ) is the point you're approximating around.
- ( \Delta x = 0.001 ) is the small change in ( x ).

First, find the derivative of ( f(x) = x^5 ): [ f'(x) = 5x^4 ]

Then evaluate ( f'(x) ) at ( x = 2 ): [ f'(2) = 5(2)^4 = 80 ]

Now, apply the formula: [ f(2.001) \approx f(2) + f'(2) \cdot 0.001 ] [ \approx 2^5 + 80 \cdot 0.001 ] [ \approx 32 + 0.08 ] [ \approx 32.08 ]

So, ( (2.001)^5 ) is approximately 32.08.

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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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