Originally Posted by
Soroban Hello, Punch!
If this is a derivative, you lost a minus-sign . . .
I assume the original function is: /$\displaystyle f(x)\;=\;\frac{\sqrt{1-x}}{\sqrt{x^2+3}} \;=\;\frac{(1-x)^{\frac{1}{2}}} {(x^2+3)^{\frac{1}{2}}} $
Then: .$\displaystyle f'(x) \;=\;\frac{(x^2+3)^{\frac{1}{2}}\cdot\frac{1}{2}(1-x)^{-\frac{1}{2}}{\color{red}(-1)} - (1-x)^{\frac{1}{2}}\cdot\frac{1}{2}(x^2+3)^{-\frac{1}{2}}(2x)}{x^2+3} $
. . . . . . . . $\displaystyle =\;\frac{\sqrt{x^2+3}\left(\dfrac{{\color{red}-1}}{2\sqrt{1-x}}\right) - \sqrt{1-x}\left(\dfrac{x}{\sqrt{x^2+3}}\right)} {x^2+3} $
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Multiply top and bottom by $\displaystyle 2\sqrt{x^2+3}\,\sqrt{1-x}$
. . and we have: .$\displaystyle \frac{-(x^2+3) - 2x(1-x)}{2(x^2+3)^{\frac{3}{2}}\sqrt{1-x}} \;=\;0$
. . which simplifies to: .$\displaystyle \frac{x^2-2x-3}{2(x^2+3)^{\frac{3}{2}}\sqrt{1-x}} \;=\;0 $
And we have: .$\displaystyle x^2 - 2x - 3 \:=\:0 \quad\Rightarrow\quad (x+1)(x-3) \:=\:0$
. . Hence: .$\displaystyle x \;=\;-1,\;3$
But because of the $\displaystyle \sqrt{1-x}$ in the function, $\displaystyle x \leq 1$
. . Therefore, the only critical value is: .$\displaystyle x \;=\;-1$