Ty Moore

Answered

2022-11-22

Interval Notation for Increasing and Decreasing Intervals of a Function

This was brought up by another student in one of my pre-calculus classes.

The graph was a simple quadratic ${x}^{2}$. The teacher stated that the graph was decreasing from $(-\mathrm{\infty},0)$, and increasing from $(0,\mathrm{\infty})$.

Why would zero not be included? i.e: decr. $(-\mathrm{\infty},0]$ and incr. $[0,\mathrm{\infty})$

This was brought up by another student in one of my pre-calculus classes.

The graph was a simple quadratic ${x}^{2}$. The teacher stated that the graph was decreasing from $(-\mathrm{\infty},0)$, and increasing from $(0,\mathrm{\infty})$.

Why would zero not be included? i.e: decr. $(-\mathrm{\infty},0]$ and incr. $[0,\mathrm{\infty})$

Answer & Explanation

reinmelk3iu

Expert

2022-11-23Added 21 answers

Explanation:

Because for f(x) to be decreasing ${f}^{\prime}(x)<0$. And for increasing ${f}^{\prime}(x)>0$. But at $x=0$, ${f}^{\prime}(x)=0$ hence it's neither decreasing nor increasing at $x=0$

Because for f(x) to be decreasing ${f}^{\prime}(x)<0$. And for increasing ${f}^{\prime}(x)>0$. But at $x=0$, ${f}^{\prime}(x)=0$ hence it's neither decreasing nor increasing at $x=0$

figoveck38

Expert

2022-11-24Added 1 answers

Explanation:

Generally the 0 is not included because the function is not decreasing (or increasing) at 0.

It would be accurate, however to say that $y={x}^{2}$ is non-increasing on the interval $(-\mathrm{\infty},0]$

Generally the 0 is not included because the function is not decreasing (or increasing) at 0.

It would be accurate, however to say that $y={x}^{2}$ is non-increasing on the interval $(-\mathrm{\infty},0]$

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