Find the minimum value $\cfrac{5-3x}{\sqrt{1-x^2}}$

Question

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 $x=\cfrac{3}{5}$.

Answer 1

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 $\,x=\dfrac ba=\dfrac35\,.$

Answer 2

Write $k = \frac{5-3x}{\sqrt{1-x^2}}$. Then we are trying to find the minimal value of $k,$ where $$ \frac{1}{k}(5-3x)=\sqrt{1-x^2}.$$ Note that the graph of the LHS is a line passing through (5/3,0). The graph of the RHS is a half-circle. The two graphs must intersect; can you find the range of $1/k$?

Answer 3

$$\frac{5-3x}{\sqrt{1-x^2}}=:y{\implies(5-3x)^2=y^2(1-x^2)\\ \implies25-30\color{#0be6a4}{x}+9\color{#0be6a4}{x^2}=y^2-y^2\color{#0be6a4}{x^2}\\ \implies\color{#808f00}{(9+y^2)}\color{#0be6a4}{x^2}\color{#9d0be6}{-30}\color{#0be6a4}{x}+\color{#850303}{(25-y^2)}=0}$$

Now, Discriminant $(\Delta =\color{#9d0be6}{b}^2-4\color{#808f00}{a}\color{#808f00}{c})$ must be positive as $x\in\mathbb{R}$,

\begin{align}&\quad\quad\quad\quad\therefore\Delta_{\color{#0be6a4}{x}} \ge0\\ &\implies900-4(9+y^2)(25-y^2)\ge0 \\&\implies y^4-16y^2\ge0 \\&\implies y\ge4\quad(\because y>0)\end{align}

Hence,$$\boxed{\min\{y\}=\min \left \{ \frac{5-3x}{\sqrt{1-x^2}} \right \}=4}$$

Calculate the value of $x$,

$$25x^2-30x+9=0{\implies(5x-3)^2=0\\\implies \boxed{x=\frac{3}5}}$$

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Generalization:

For, $$\frac{a-bx}{\sqrt{1-x^2}}=:z$$

\begin{align}&\quad\quad\quad\quad\therefore\Delta \ge0\\ &\implies 4a^2b^2-4(b^2+z^2)(a^2-z^2)\ge0 \\&\implies z^4-(a^2-b^2)z^2\ge0 \\&\implies z\ge\sqrt{a^2-b^2}\quad(\because z>0)\end{align}

Hence,$$\boxed{\min\{z\}=\min \left \{ \frac{5-3x}{\sqrt{1-x^2}} \right \}=\sqrt{a^2-b^2}}$$