# How do you use linear approximation to the square root function to estimate square roots #sqrt 3.60#?

compared to exact value of

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To use linear approximation to estimate the square root of 3.60, you start with a known value close to 3.60. Let's take ( x = 3.6 ). Then, we find the square root of this known value ( f(x) = \sqrt{x} ).

Next, we find the derivative of ( f(x) ) at ( x = 3.6 ), denoted as ( f'(x) ).

After finding ( f(x) ) and ( f'(x) ), we use the linear approximation formula:

[ f(a + \Delta x) ≈ f(a) + f'(a) * \Delta x ]

where ( a ) is the known value (in this case, 3.6), ( f(a) ) is the known square root of that value, and ( \Delta x ) is the difference between the value you're estimating and the known value (in this case, ( \Delta x = 3.60 - 3.6 = 0.0 )).

Plugging in the values, you can approximate ( \sqrt{3.60} ) using linear approximation.

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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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