How do you find the slope of the secant lines of $f (x) = \sqrt{x}$ $f(x) = sqrtx$ at P(4,2)?

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How do you find the slope of the secant lines of $f (x) = \sqrt{x}$ $f(x) = sqrtx$ at P(4,2)?

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Answer 1 · 2017-03-16T03:22:08

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Explanation:

$\frac{f (4 + h) - f (4)}{(4 + h) - (4)} = \frac{\sqrt{4 + h} - 2}{h}$ $(f(4+h)-f(4))/((4+h)-(4)) = (sqrt(4+h)-2)/h$

$= \frac{(\sqrt{4 + h} - 2)}{h} \frac{(\sqrt{4 + h} + 2)}{(\sqrt{4 + h} + 2)}$ $= ((sqrt(4+h)-2))/h ((sqrt(4+h)+2))/((sqrt(4+h) +2))$

$= \frac{(4 + h) - 4}{h (\sqrt{4 + h} + 2)}$ $= ((4+h)-4)/(h(sqrt(4+h)+2)$

$= \frac{1}{(\sqrt{4 + h} + 2)}$ $= 1/((sqrt(4+h)+2)$

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How do you find the slope of the secant lines of $f (x) = \sqrt{x}$ $f(x) = sqrtx$ at P(4,2)?

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Explanation:

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How do you find the slope of the secant lines of $f (x) = \sqrt{x}$ $f(x) = sqrtx$ at P(4,2)?