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

Dec 11, 2015

Increasing.

#### Explanation:

Find the first derivative of the function.

If $f ' \left(5\right) < 0$, then $f \left(x\right)$ is decreasing when $x = 5$.
If $f ' \left(5\right) > 0$, then $f \left(x\right)$ is increasing when $x = 5$.

To find $f ' \left(x\right)$, use the quotient rule.

$f ' \left(x\right) = \frac{\sqrt{x + 3} \frac{d}{\mathrm{dx}} \left[x\right] - x \frac{d}{\mathrm{dx}} \left[\sqrt{x + 3}\right]}{\sqrt{x + 3}} ^ 2$

Find each derivative separately.

$\frac{d}{\mathrm{dx}} \left[x\right] = 1$

The following derivative requires the chain rule.

$\frac{d}{\mathrm{dx}} \left[\sqrt{x + 3}\right] = \frac{1}{2} {\left(x + 3\right)}^{- \frac{1}{2}} {\overbrace{\frac{d}{\mathrm{dx}} \left[x + 3\right]}}^{= 1} = \frac{1}{2 \sqrt{x + 3}}$

Plug the derivatives back in.

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

Multiply everything by $2 \sqrt{x + 3}$ to clear out the denominator of the internal fraction.

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

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

Now, find $f ' \left(5\right)$.

$f ' \left(5\right) = \frac{5 + 6}{2 {\left(5 + 3\right)}^{\frac{3}{2}}} = \frac{11}{2 {\left(8\right)}^{\frac{3}{2}}}$

We could determine the exact value of $f ' \left(5\right)$, but it should be clear that the answer will be positive.

Thus, $f ' \left(x\right)$ is increasing when $x = 5$.