Question

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

Answer 1

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^'(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]}