Question

Over what intervals is `f(x)=(x-1)^2-x^3+x` increasing and decreasing?

Answer 1

`f(x)` is decreasing on `x inRR` (on the interval `-oo < x < oo`).

Explanation:

We will have to differentiate the function:

• If `f'>0`, then `f` is increasing.
• If `f'<0`, then `f` is decreasing.

First, simplify `f` by expanding `(x-1)^2` and then combining like terms.

`f(x)=x^2-2x+1-x^3+x`

`f(x)=-x^3+x^2-x+1`

Now, find `f'` through the power rule.

`f'(x)=-3x^2+2x-1`

In order to analyze when `f'` is positive or negative, we must find when it could change sign, which is when `f'=0`. When we analyze

`-3x^2+2x-1=0`

We see that the polynomial has a negative discriminant, which means the that `f'` never equals `0`. Since the function is continuous, and that the function `f'(x)=-3x^2+2x-1` will always be `<0`, since it is a downwards-facing parabola with no real roots, we can determine that the graph of `f` is always decreasing.

The graph of `f`, which is always decreasing:

graph{(x-1)^2-x^3+x [-13.41, 15.07, -6.32, 7.92]}

The graph of `f'`, which is always negative:

graph{-3x^2+2x-1 [-20.13, 20.42, -12.77, 7.5]}