# Is f(x)=x-1/sqrt(x^3-3x) increasing or decreasing at x=2?

Jun 15, 2016

Increasing.

#### Explanation:

We will have to find the sign of the function's derivative at $x = 2$. This will tell us whether the function is increasing or decreasing that that point:

• If ${f}^{'} \left(2\right) < 0$, then $f \left(x\right)$ is decreasing at $x = 2$.
• If ${f}^{'} \left(2\right) > 0$, then $f \left(x\right)$ is increasing at $x = 2$.

So, we need to differentiate the function. Rewrite it first using fractional exponents:

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

The initial term $x$ is easily differentiated as $1$. The second term will require the chain rule. Note that:

$\frac{d}{\mathrm{dx}} \left({u}^{- \frac{1}{2}}\right) = - \frac{1}{2} {u}^{- \frac{3}{2}} \cdot {u}^{'}$

So, we see that:

${f}^{'} \left(x\right) = 1 - \left(- \frac{1}{2}\right) {\left({x}^{3} - 3 x\right)}^{- \frac{3}{2}} \cdot \frac{d}{\mathrm{dx}} \left({x}^{3} - 3 x\right)$

Simplifying further, and differentiating the inside function with the power rule:

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

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

So, we want to determine the derivative's sign at $x = 2$:

f^'(2)=1+(3(2^2)-3)/(2(2^3-2(3))^(3/2))=1+(12-3)/(2(8-6)^(3/2))=1+9/(2(2^(3/2))

Without calculating this number's decimal value, we can tell it will be positive. Since ${f}^{'} \left(2\right) > 0$, we know that $f \left(x\right)$ is increasing at $x = 2$.

We can check a graph of $f \left(x\right)$:

graph{x-(x^3-3x)^(-1/2) [-6.1, 11.68, -4.615, 4.275]}