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

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^'(2)<0#, then #f(x)# is decreasing at #x=2#.
• If #f^'(2)>0#, then #f(x)# is increasing at #x=2#.

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

#f(x)=x-(x^3-3x)^(-1/2)#

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

#d/dx(u^(-1/2))=-1/2u^(-3/2)*u^'#

So, we see that:

#f^'(x)=1-(-1/2)(x^3-3x)^(-3/2)*d/dx(x^3-3x)#

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

#f^'(x)=1+1/2(x^3-3x)^(-3/2)*(3x^2-3)#

#f^'(x)=1+(3x^2-3)/(2(x^3-3x)^(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^'(2)>0#, we know that #f(x)# is increasing at #x=2#.

We can check a graph of #f(x)#:

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