Is $f (x) = \frac{x - 3}{\sqrt{x + 3}}$ $f(x)=(x-3)/sqrt(x+3)$ increasing or decreasing at $x = 3$ $x=3$ ?

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Is $f (x) = \frac{x - 3}{\sqrt{x + 3}}$ $f(x)=(x-3)/sqrt(x+3)$ increasing or decreasing at $x = 3$ $x=3$ ?

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Answer 1 · 2016-11-24T21:55:15

You have to look at the sign of the derivative of the function for $x = 3$ $x=3$

Explanation:

$f (x) = \frac{x - 3}{\sqrt{x + 3}} = (x - 3) {(x + 3)}^{- \frac{1}{2}}$ $f(x) = (x-3)/sqrt(x+3) = (x-3)(x+3)^(-1/2)$

$\frac{d f}{d x} = {(x + 3)}^{- \frac{1}{2}} - \frac{1}{2} (x - 3) {(x + 3)}^{- \frac{3}{2}}$ $(df)/(dx) = (x+3)^(-1/2) -1/2 (x-3)(x+3)^(-3/2)$

$\frac{d f}{d x} ∣_{x = 3} = \frac{1}{\sqrt{6}} > 0$ $(df)/(dx) |_(x=3) = 1/sqrt(6) > 0$

So the function is increasing.

On the other hand it is easy to see that $x = 3$ $x=3$ is a zero of $f (x)$ $f(x)$ and as the denominator is always positive and the numerator is negative for $x < 3$ $x<3$ and positive for $x > 3$ $x>3$ the function must be increasing.

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Is $f (x) = \frac{x - 3}{\sqrt{x + 3}}$ $f(x)=(x-3)/sqrt(x+3)$ increasing or decreasing at $x = 3$ $x=3$ ?

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Is $f (x) = \frac{x - 3}{\sqrt{x + 3}}$ $f(x)=(x-3)/sqrt(x+3)$ increasing or decreasing at $x = 3$ $x=3$ ?