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

Increasing.

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

So, we see that:

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

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

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To determine if ( f(x) = \frac{x - 1}{\sqrt{x^3 - 3x}} ) is increasing or decreasing at ( x = 2 ), we need to examine the sign of its derivative at that point.

First, find the derivative of ( f(x) ) using the quotient rule:

[ f'(x) = \frac{(1)(\sqrt{x^3 - 3x}) - (x - 1)\left(\frac{1}{2}(x^3 - 3x)^{-1/2}(3x^2 - 3)\right)}{x^3 - 3x} ]

Simplify the expression:

[ f'(x) = \frac{\sqrt{x^3 - 3x} - \frac{1}{2}(x - 1)(x^3 - 3x)^{-1/2}(3x^2 - 3)}{x^3 - 3x} ]

Now evaluate ( f'(2) ):

[ f'(2) = \frac{\sqrt{2^3 - 3(2)} - \frac{1}{2}(2 - 1)(2^3 - 3(2))^{-1/2}(3(2)^2 - 3)}{2^3 - 3(2)} ]

[ f'(2) = \frac{\sqrt{8 - 6} - \frac{1}{2}(1)(8 - 6)^{-1/2}(12 - 3)}{8 - 6} ]

[ f'(2) = \frac{\sqrt{2} - \frac{1}{2}(2)(2)^{-1/2}(9)}{2} ]

[ f'(2) = \frac{\sqrt{2} - \frac{9}{\sqrt{2}}}{2} ]

[ f'(2) = \frac{\sqrt{2} - \frac{9\sqrt{2}}{2}}{2} ]

[ f'(2) = \frac{2\sqrt{2} - 9\sqrt{2}}{4} ]

[ f'(2) = \frac{-7\sqrt{2}}{4} ]

Since ( f'(2) ) is negative, ( f(x) ) is decreasing at ( x = 2 ).

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