# Is f(x)=x-sqrt(x^3-3x) increasing or decreasing at x=2?

Feb 19, 2016

Decreasing.

#### Explanation:

Let's find the first derivative of the function $f$, at $x = 2$ using $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$, thereafter the chain rule and then inserting the desired value of $x$.

$f \left(x\right) = x - \sqrt{{x}^{3} - 3 x}$
${f}^{'} \left(x\right) = 1 - \frac{1}{2} {\left({x}^{3} - 3 x\right)}^{- \frac{1}{2}} \times \left(3 {x}^{2} - 3\right)$
Now
${f}^{'} \left(2\right) = 1 - \frac{1}{2} {\left({2}^{3} - 3 \times 2\right)}^{- \frac{1}{2}} \times \left(3 \times {2}^{2} - 3\right)$
$= 1 - \frac{1}{2} {\left(8 - 6\right)}^{- \frac{1}{2}} \times \left(12 - 3\right)$
$= 1 - \frac{1}{2} \times \frac{1}{\sqrt{2}} \times 9$
$\approx - 2.18$

Since slope and derivative are synonymous, we observe that slope of the function at $x = 2$ is negative. Hence the function is decreasing at this value of $x$

It can be verified by drawing graph of the function using the graphing tool.
graph{y=x-sqrt(x^3-3x) [-3.437, 6.56, -2.76, 2.24]}