Without using the derivative, find the minimum value and find the $x$ at which it is achieved
$$\min \left \{ \frac{5-3x}{\sqrt{1-x^2}} \right \}$$
I tried to substitute $x=\cos t$ and got $\cfrac{5-3\cos t}{|\sin t|}$ and there is no sense in the module. Without it, you can also find the point of minimum $\cfrac{5-3\cos t}{\sin t}$. I don't really understand how to find the smallest value of the new function and find the point $t=\arccos \cfrac{3}{5}$ to do the inverse substitution and get $x=\cfrac{3}{5}$.
Best Answer
Note that
$(a-bx)^2-(b-ax)^2=(a^2-b^2)(1-x^2)\;.$
Moreover, if $\,a>b\geqslant0\,,\,$ it results that
$a-bx\geqslant0\quad$ for any $\;x\in\,]\!-\!\infty,1]\supsetneq(-1,1)\,.$
Consequently, if $\,a>b\geqslant0\,,\,$ it results that
$\!\!\!\!\!\!\begin{align}\dfrac{a-bx}{\sqrt{1-x^2}}&\!=\!\sqrt{\dfrac{(a\!-\!bx)^2}{1-x^2}}\!=\!\sqrt{\dfrac{(a\!-\!bx)^2\!\!-\!(b\!-\!ax)^2\!\!+\!(b\!-\!ax)^2}{1-x^2}}\!=\end{align}$
$\begin{align}=\sqrt{\dfrac{(a^2\!\!-\!b^2)(1\!-\!x^2)\!+\!(b\!-\!ax)^2}{1-x^2}}&=\sqrt{a^2\!-\!b^2\!+\!\dfrac{(b\!-\!ax)^2}{1-x^2}}\geqslant\end{align}$
$\geqslant\sqrt{a^2\!-\!b^2}\quad$ for any $\,x\in(-1,1)\,.$
Hence, if $\,a>b\geqslant0\,,\,$ the minimum value of $\;\dfrac{a-bx}{\sqrt{1-x^2}}\;$ is $\;\sqrt{a^2-b^2}\;$ and it is achieved for $\,x=\dfrac ba\in[0,1[\,.$
In particular, if $\,a=5\,$ and $\,b=3\,,\,$ the minimum value is $\;\sqrt{a^2-b^2}=4\;$ and it is achieved for $\,x=\dfrac ba=\dfrac35\,.$