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

Question

Luke Alvarez · Accepted Answer

See below

#(f(4+h)-f(4))/((4+h)-(4)) = (sqrt(4+h)-2)/h# # = ((sqrt(4+h)-2))/h ((sqrt(4+h)+2))/((sqrt(4+h) +2))# # = ((4+h)-4)/(h(sqrt(4+h)+2)# # = 1/((sqrt(4+h)+2)#

Charles Arocha · Answer

undefined 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} $.