Monotonic function

  • #1
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Homework Statement


function [tex]y(x) = x^2 + x + 1[/tex]

The Attempt at a Solution


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

Answers and Replies

  • #2
lanedance
Homework Helper
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i'd say two open intervals
f(x) is decreasing for [itex]x \in \left(- \infty, -\frac{1}{2}\right) [/itex]
f(x) is increasing for [itex]x \in \left(-\frac{1}{2}\right, \infty \right) [/itex]

and neither at [itex]x = -\frac{1}{2} [/itex]
 
  • #3
SammyS
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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] .
 

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