Why did he write $f_\min=3/2$?

Question

I don't have much reputation to comment everywhere as as I'm new here.

Could some explain me what and why this line is being written in this anser to "Find minimum value of $f(b)$ where $f(b)$ denotes the maximum value of $g(x)$ where $g(x)=\left|\sin x+\frac{2}{3+\sin x}+b\right|$." I didn't really understand despite reading it several times.

If $b\leq -{3\over 2}$ then $$h(t) = -t-{2\over t}-b+3 \leq -b = f(b) \implies \min f(b) = {3\over 2}$$

If $-{3\over 2}<b<0$ then $$f(b)= \max \{h(2),h(4)\} = \max\left\{-b,{3\over 2}+b \right\}.$$

Answer 1

It was stated in the answer that $$f(b)=\max_{2\le t\le 4}h(t),\quad\text{where}\quad h(t)=\left|t+\frac2t-3+b\right|$$ and that when $t$ increases from $2$ to $4,$ $t+\frac2t$ increases from $3$ to $\frac92.$ Therefore, $t+\frac2t-3$ increases from $0$ to $\frac32,$ and $$f(b)=\max_{0\le s\le\frac32}|s+b|.$$ If $b\le-\frac32$ then $\forall s\le\frac32\quad s+b\le0$ and $$f(b)=\max_{0\le s\le\frac32}(-s-b)=-b.$$ If $-\frac32\le b\le0$ then $s+b$ is $\le0$ for $s\in[0,-b]$ and $\ge0$ for $s\in\left[-b,\frac32\right],$ so $$f(b)=\max\left(\max_{0\le s\le-b}(-s-b),\max_{-b\le s\le\frac32}(s+b)\right)=\max\left(-b,\frac32+b\right).$$ Note that the value $-b$ corresponds in both cases to $s=0$ i.e. $t=2,$ and the value $\frac32+b$ (in the second case) to $s=\frac32$ i.e. $t=4:$ $$-b=h(0),\qquad\frac32+b=h(4).$$

Similarly, they found that if $b\ge0$ then $f(b)=\frac32+b,$ so that finally, $$\min_{b\in\Bbb R}f(b)=\min\left(\min_{b\le-\frac32}-b,\min_{-\frac32\le b\le0}\max\left(-b,\frac32+b\right),\min_{b\ge0}\frac32+b\right)$$ $$=\min\left(\frac32,\frac34,\frac32\right)=\frac34.$$

Answer 2

The answer states the bounds for $t$ between $2 \leq t \leq 4$. If $t=4$ then $$h(t)=t+\frac{2}{t}+b-3 = 4+\frac{1}{2}+b-3=\frac{3}{2}+b$$

Which means the minimum can be $\frac{3}{2}$ since the condition $b=0$ is allowed.

Does this help?

Answer 3

As in the answer defined we have $h(t)=|t-3+\frac{2}{t}+b|$. The absolute value is annoying therefore we distinct 3 cases.

For $b\geq0$ we have $t-3+\frac{2}{t}+b\geq0$.

For $b \leq -1.5$ we have $t-3+\frac{2}{t}+b\leq0$.

In both cases you can resolve the absolute value, maximize $h(t)$ and eventually minimize $f(b)$.

The question is what happens for $-1.5\leq b \leq 0$? We know that the absolute value takes its maximum on its boundaries therefore we get $\max h=\max(h(2),h(4))$ an from there on we can solve the $\min f(b)$.