<s>[INST] Monotonicity of function at a point 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 $b\in (-2,-1)\cup (1,+\infty)$ 
 [/INST] $f$ 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 $(1,\infty$)).
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