# Is f(x)=(x-3)/sqrt(x+3)  increasing or decreasing at x=5 ?

##### 1 Answer
Nov 22, 2016

We start by differentiating.

Let $y = {u}^{\frac{1}{2}}$ and $u = x + 3$. By the power rule, $y ' = \frac{1}{2} {u}^{- \frac{1}{2}} = \frac{1}{2 {u}^{\frac{1}{2}}}$ and $u ' = 1$

By the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 {u}^{\frac{1}{2}}} \times 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 {\left(x + 3\right)}^{\frac{1}{2}}}$

For $f ' \left(x\right)$, we find the derivative by the quotient rule:

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

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

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

$f ' \left(x\right) = \frac{\frac{2 \left(x + 3\right) - x + 3}{2 {\left(x + 3\right)}^{\frac{1}{2}}}}{x + 3}$

$f ' \left(x\right) = \frac{\frac{2 x + 6 - x + 3}{2 {\left(x + 3\right)}^{\frac{1}{2}}}}{x + 3}$

$f ' \left(x\right) = \frac{x + 9}{2 {\left(x + 3\right)}^{\frac{1}{2}} x + 3}$

$f ' \left(x\right) = \frac{x + 9}{2 {\left(x + 3\right)}^{\frac{3}{2}}}$

Now, if the function is increasing at the point $x = a$, then $f ' \left(a\right) > 0$. If the function is decreasing at $x = a$, then $f ' \left(a\right) < 0$.

$f ' \left(5\right) = \frac{5 + 9}{2 {\left(5 + 3\right)}^{\frac{3}{2}}}$

$f ' \left(5\right) = \frac{14}{2 \sqrt{512}}$

$f ' \left(5\right) = \frac{14}{2 \left(16\right) \sqrt{2}}$

f'(5) = 14/(32sqrt(2)

$f ' \left(5\right) = \frac{7}{16 \sqrt{2}}$

What matters is that the above value is positive. Hence, $f \left(x\right)$ is increasing at $x = 5$.

Hopefully this helps!