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I know that ,based on definitions, in the interval $(a,b)$

For increasing functions, $f(a) \leq f(b)$ and $f'(x) \geq 0$

For strictly increasing functions, $f(a)<f(b)$ and $f'(x)>0$

However,

For example, I am given a question:

Given that $y=4x^3 +3x^2 -6x -7$, find the range of values of $x$ for which $y$ is decreasing.

For $y$ to be decreasing, $\frac{dy}{dx} \leq 0$

So, $-1\leq x\leq\frac12$ right?

But for strictly decreasing, $-1<x<\frac12$ right?

Then for increasing, it should be $x \leq -1$ and $x\geq \frac12$

For strictly increasing, it should be $x< -1$ and also $x \geq \frac12$ (I heard it can start from $\frac{dy}{dx}=0$ )

Basically, the problem is what type of inequality sign do I use for which case?

Yes, I did see the other posts on this topic and some quora posts. But I'm still lost.

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  • $\begingroup$ Strictly increasing does not require the function to be differentiable, not that the derivative, if it exists, be strictly positive everywhere. The function $x \mapsto x^3$ is strictly increasing but the derivative is zero at $x=0$. $\endgroup$
    – copper.hat
    Oct 27, 2017 at 22:37
  • $\begingroup$ A function is increasing if for all $a < b$ implies that $f(a) \le f(b)$, and strictly increasing if for all $a < b$ implies that $f(a) < f(b)$. These are the definitions you should have in mind. A consequence of these definitions is that (strictly) increasing functions will have nonnegative derivatives (when they have derivatives at all). However, even a strictly increasing function can have points with zero derivative (e.g. $x \mapsto x^3$ is strictly increasing, and has derivative zero at $x=0$). $\endgroup$
    – Xander Henderson ♦
    Oct 27, 2017 at 22:37
  • $\begingroup$ However if the question asks to find the range at which the function is strictly increasing then what inequality sign do I use? And do I have to include the stationary points or exclude? $\endgroup$
    – Arc Neoepi
    Oct 27, 2017 at 22:40
  • $\begingroup$ If a continuous function is strictly increasing on an interval $(a,b)$ then it is also strictly increasing on the interval $[a,b]$. In your example, the function is strictly increasing on $(-\infty,-1]$ and on $\left[\frac{1}{2},\infty\right)$ and strictly decreasing on the interval $\left[-1,\frac{1}{2}\right]$. [Assuming, of course, that $a$ and $b$ are in the domain.] $\endgroup$
    – John Wayland Bales
    Oct 27, 2017 at 22:45
  • $\begingroup$ @JohnWaylandBales So, the example that I have given, the function is strictly increasing and increasing in the same interval right? The same should go for the interval for strictly decreasing and decreasing too right? $\endgroup$
    – Arc Neoepi
    Oct 27, 2017 at 22:50

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