Monotonic function

[player1_1_1]

Homework Statement

function $$y(x) = x^2 + x + 1$$

The Attempt at a Solution

I count derivative: $$f^{\prime} (x) = 2x + 1$$ and now $$f^{\prime (x) = 0$$ when $$x=-\frac{1}{2}$$ and how to describe monotonic now? $$f(x)$$ is decreasing for $$x \in \left(- \infty; -\frac{1}{2}\right]$$ or $$x \in \left(- \infty; -\frac{1}{2}\right)$$? open or closed interval? and now increasing for what $$x$$?

 

[lanedance]

i'd say two open intervals
f(x) is decreasing for $x \in \left(- \infty, -\frac{1}{2}\right)$
f(x) is increasing for $x \in \left(-\frac{1}{2}\right, \infty \right)$

and neither at $x = -\frac{1}{2}$

 

[SammyS]

I disagree.

f(x) has an absolute minimum of ‒3/4 at x = ‒1/2.

f(x)>f(‒1/2) for x > ‒1/2, so f(x) is monotonic increasing on [‒1/2, +∞) .

Similarly, f(x) is monotonic decreasing on (‒∞ , ‒1/2] .