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Let $\displaystyle f(b) = \max\bigg|\sin x+\frac{2}{3+\sin x}+b\bigg|\forall x\in \mathbb{R}.$ Then $\min (f(b))\forall b \in \mathbb{R}$

$\bf{My\; Try::}$ Let $\displaystyle y=\frac{\sin^2 x+3\sin x+2}{3+\sin x}\;,$ Put $\sin x = z\;,$ Then $|z|\leq 1$

and $\displaystyle y = \frac{z^2+3z+2}{3+z} = \frac{(z+1)(z+2)}{(z+3)}\;,$ put $z+3=u$

So $\displaystyle y = \frac{(u-1)(u-2)}{u} = u+\frac{2}{u}-3\; u \in [2,4]$

which is strictly Increasing function in $[2,4]$

So $y \in \left[0,\frac{3}{2}\right]$

So we get $\displaystyle f(b) = \max \left|b,\frac{3}{2}+b\right|=\frac{3}{2}+b$

Now how can i solve it , Help required, Thanks

  • 0
    does the global min exist? – 2017-01-25
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    Answer given as $\displaystyle \min(f(b)) = \frac{3}{4}\forall b \in \mathbb{R}$ at $\displaystyle b=-\frac{3}{4}$ – 2017-01-25
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    min exists. i forgot that $b$ is inside the absolute value sign. – 2017-01-25

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Your last expression for $f(b)$ is not correct. $\sin x+\frac{2}{3+\sin x}$ varies between $0$ and $3/2$, therefore $$ f(b) = \max \{ \lvert b \rvert , \lvert b + \frac 32 \rvert \} $$ Now draw the graph of that function to see that the minimum is attained at $b = -\frac 34$.