Vtyjkyuk

The question says :

Let
$$f(x)=begincases
-x^3+fracb^3-b^2+b-1b^2+3b+2 &:0le xlt1\ 2x-3 &:1le xle3endcases$$. 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 $bin (-2,-1)cup (1,+infty)$

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 $f(1)$.

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.

$f$ 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 $(1,infty$)).