Monotonic Function: Derivative and Interval Analysis for f(x) = x^2 + x + 1

[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] .