Question

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

Answer 1

Decreasing.

Explanation:

Let's find the first derivative of the function `f`, at `x=2` using `d/dxx^n=nx^(n-1)`, thereafter the chain rule and then inserting the desired value of `x`.

`f(x)=x-sqrt(x^3-3x)`
`f^'(x)=1-1/2(x^3-3x)^(-1/2)xx(3x^2-3)`
Now
`f^'(2)=1-1/2(2^3-3xx2)^(-1/2)xx(3xx2^2-3)`
`=1-1/2(8-6)^(-1/2)xx(12-3)`
`=1-1/2xx 1/sqrt 2xx9`
`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]}