# How do you find the slope of the secant lines of #f(x) = sqrtx# at P(4,2)?

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To find the slope of the secant line of ( f(x) = \sqrt{x} ) at ( P(4,2) ), we first find the coordinates of another point on the secant line. Let's choose a point ( Q ) with ( x )-coordinate ( x ) on the curve. The coordinates of ( Q ) will be ( (x, \sqrt{x}) ).

The slope of the secant line passing through ( P(4,2) ) and ( Q(x, \sqrt{x}) ) is given by:

[ \text{Slope of secant} = \frac{\text{Change in } y}{\text{Change in } x} = \frac{\sqrt{x} - 2}{x - 4} ]

To find the slope at the specific point ( P(4,2) ), we evaluate this expression as ( x ) approaches 4:

[ \text{Slope at } P(4,2) = \lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4} ]

Using the limit laws and direct substitution, we get:

[ \text{Slope at } P(4,2) = \frac{\sqrt{4} - 2}{4 - 4} = \frac{0}{0} ]

This form indicates that we need to further simplify the expression. We can do this by rationalizing the numerator:

[ \text{Slope at } P(4,2) = \lim_{x \to 4} \frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{(x - 4)(\sqrt{x} + 2)} ]

[ = \lim_{x \to 4} \frac{x - 4}{(x - 4)(\sqrt{x} + 2)} ]

[ = \lim_{x \to 4} \frac{1}{\sqrt{x} + 2} ]

Now, substitute ( x = 4 ) into the simplified expression:

[ \text{Slope at } P(4,2) = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4} ]

Therefore, the slope of the secant line of ( f(x) = \sqrt{x} ) at ( P(4,2) ) is ( \frac{1}{4} ).

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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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