Compute Δy and dy for x = 16 and dx = Δx = 1. (Round the answers to three decimal places.)? y = √x
what is Δy
what is Δy
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Given ( y = \sqrt{x} ), compute ( \Delta y ) and ( dy ) for ( x = 16 ) and ( \Delta x = dx = 1 ).
[ \Delta y = y(x + \Delta x) - y(x) ]
[ dy = y'(x) \cdot dx ]
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Calculate ( y(x) ) when ( x = 16 ): [ y(16) = \sqrt{16} = 4 ]
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Calculate ( y(x + \Delta x) ) when ( x = 16 ) and ( \Delta x = 1 ): [ y(16 + 1) = y(17) = \sqrt{17} ]
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Calculate ( \Delta y ): [ \Delta y = \sqrt{17} - 4 ]
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Calculate ( y'(x) ): [ y'(x) = \frac{1}{2\sqrt{x}} ]
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Calculate ( y'(16) ): [ y'(16) = \frac{1}{2\sqrt{16}} = \frac{1}{8} ]
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Calculate ( dy ): [ dy = \frac{1}{8} \cdot 1 = \frac{1}{8} ]
So, ( \Delta y \approx \sqrt{17} - 4 ) and ( dy = \frac{1}{8} ).
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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.
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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