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The question says :

Let $$f(x)=\begin{cases} -x^3+\frac{b^3-b^2+b-1}{b^2+3b+2} &:0\le x\lt1\\ 2x-3 &:1\le x\le3\end{cases}$$. Find all possible values of b such that f(x)has the smallest value at $x=1$.

Since this question was an example question, the solution said,

The Limiting value of f(x) from the left of $x=1$ should be either greater or equal to the value of the function at $x=1$.

My question is for $x=1$ to have the smallest possible value, shouldn't the limiting value be Less than or equal to the function at $x=1$?

The answer is $b\in (-2,-1)\cup (1,+\infty)$

delusional.existence
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alburrito
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  • "Monotonicity at a point"? What is that? – DonAntonio Aug 21 '18 at 08:15
  • Try to sketch a picture of your situation. – MSDG Aug 21 '18 at 08:16
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    @DonAntonio: This is a fairly well known notion, although it's often not mentioned in elementary calculus texts. It's in Spivak's calculus book, however. See Can a function be increasing at a point? for a general discussion, and see Brown/Darji/Larsen's Nowhere monotone functions and functions of nonmonotonic type for a relatively accessible paper involving this notion. – Dave L. Renfro Aug 21 '18 at 08:28
  • @Dave L. Renfro I see what you mean, but while I can understand the increasing/decreasing/constant nature of a function at say, $x_0$, in terms of the sign of $f(x_0)$; I fail to understand what is meant by 'monotocity at a point'. Doesn't monotocity mean -does not change the sign of ts derivative, had only one nature (Inc./Dec.). But the derivative can have only one value at a point. – user0 Aug 21 '18 at 08:44
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    @DaveL.Renfro "Function increasing at a point" I've heard, though I think it is not very common. Monotonicity is what sounded weird. – DonAntonio Aug 21 '18 at 08:56
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    @DevashishKaushik: Monotonicity at $x$ means "non-decreasing at $x$ OR non-increasing at $x.$ Also, this notion, as well as montonicity on an interval, monotonicity on a set (that doesn't have to be an interval), etc. are defined independently of the idea of a derivative, and in fact make sense for nowhere differentiable functions. You're conflating theorems that relate derivatives to monotonicity behavior with definitions. – Dave L. Renfro Aug 21 '18 at 08:58
  • @DonAntonio: Yes, it also sounds slightly strange to me. I suspect this is because whenever these precise notions are used (in more advanced literature), people are concerned with the extra precision in stating something about pointwise increase, which winds up also being true for pointwise decrease, as compared to stating something about pointwise monotonicity. For example, it's stronger to assert the existence of a function that is increasing at each point of a set of positive measure than to assert the existence of a function that is monotone at each point of a set of positive measure. – Dave L. Renfro Aug 21 '18 at 09:06
  • @Dave L. Renfro OK, I get your point about monotocity at $x$. However, I don't understand why we are defining it this way. To refine what I said in the previous comment, doesn't monotocity on an interval usually mean that the function does not change its increasing or decreasing nature inside the interval (equivalent to the tone/sign of the derivative whenever it is defined). Or at least, that's the only way I have seen it used before this. – user0 Aug 21 '18 at 09:07
  • DevashishKaushik: I'm not quite sure what you're asking, but I believe the confusion has to do with these two notions, which would have to be proved to be the same: (1) $f$ is increasing at each point of the open interval $I;$ (2) $f$ is increasing on the open interval $I.$ Of course, if we say "monotone at each point in a set", then the increase/decrease distinction can change from point to point, but not if we way "monotone on the set", so for "monotone" the link between the local (at a point) and global (on an interval) notions is not very strong. – Dave L. Renfro Aug 21 '18 at 09:16
  • For some more discussion of local vs global monotonicity notions, see my 25 April 2000 ap-calculus post archived at Math Forum. – Dave L. Renfro Aug 21 '18 at 09:17

3 Answers3

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No, you are looking at it the wrong way.

The question asked for a condition so that $ f(x) $ attains it's smallest value ( has a global minimum) at $x=1$). This means that we want $ f(1) $ to be the smallest value that $f$ attains.

Clearly, the limiting value should be greater and not less than the value of the function.

user0
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Notice that $f$ is monotonically decreasing on $[0,1)$ and monotonically increasing on $[1,3]$. Now, if you want $f$ to attain minimum value at $1$, then $f(x) \geq f(1) \, \forall x \in [0,1)$. Hence, the left limit would be greater that or equal to $f(1)$.

Sntn
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$f$ is decreasing on $[0,1)$ and increasing on $[1,3]$. Hence $f(1)$ is the minimum iff $f(1) \geq f(1-)$ which means $-1 \geq -1+\frac {(1+b^{2})(b-1)} {(1+b)(2+b)}$ or $\frac {(b-1)} {(1+b)(2+b)}\leq 0$ . It is easy to find values of $b$ from this. (The function is not defined for $b=-1$ and $b=-2$ so consider $b$ in $(-\infty, -2),(-2,-1)$ and $(1,\infty$)).

Kavi Rama Murthy
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