# Algebra; Functions & Graphs

Given:

Why is -2 non-inclusive?

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The derivative of the function (the slope) is 0 at x = -2.

The definition of an increasing function requires that the slope be strictly positive. :)

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I presume you mean strictly increasing and are talking about continuously differentiable functions. Then if the slope is positive, it's true that $f$ is strictly increasing, but not the converse; consider $f(x)=x^3$. – Michael Hoppe Feb 28 '14 at 13:32
This is not correct. – Mark McClure Feb 28 '14 at 14:16
@MarkMcClure Do you mind to explain my error? – Michael Hoppe Feb 28 '14 at 14:25
@MichaelHoppe I mean this answer is incorrect - yours is the correct one! I even upvoted it! – Mark McClure Feb 28 '14 at 14:35
@MarkMcClure thanks for clarification. – Michael Hoppe Feb 28 '14 at 14:39

The function is strictly increasing on any interval $J\subset(-\infty,-2]$, so the suggested answer is true since $(-\infty,-2)\subset(-\infty,-2]$.

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This is correct. – Mark McClure Feb 28 '14 at 14:15