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The expression $\sqrt{-x^{2}+6 x-5}$ has a

(A) maximum value of 2
(B) minimum value of 2
(C) maximum value of 3
(D) $\quad$ minimum value of 3
in Mathematics by Platinum (119,120 points) | 68 views

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Answer:

D- a minimum value of 3

 

test

 

 

Explanation:

\(\sqrt{-x^{2}+6 x-5}\)

Simplify/rewrite:
\[
\sqrt{-(x-5)(x-1)}
\]

 

Roots/zeros found at:
\[
\begin{aligned}
&x=5 \\
&x=1
\end{aligned}
\]

First Derivative

 

\begin{aligned}
&\frac{\mathrm{d}}{\mathrm{d} x}[\sqrt{-(x-5)(x-1)}]\\
&=\frac{1}{2}(-(x-5)(x-1))^{\frac{1}{2}-1} \cdot \frac{\mathrm{d}}{\mathrm{d} x}[-(x-5)(x-1)]\\
&=\frac{-\frac{\mathrm{d}}{\mathrm{d} x}[(x-5)(x-1)]}{2 \sqrt{-(x-5)(x-1)}}\\
&=-\frac{\frac{\mathrm{d}}{\mathrm{d} x}[x-5] \cdot(x-1)+(x-5) \cdot \frac{\mathrm{d}}{\mathrm{d} x}[x-1]}{2 \sqrt{-(x-5)(x-1)}}\\
&=-\frac{\left(\frac{\mathrm{d}}{\mathrm{d} x}[x]+\frac{\mathrm{d}}{\mathrm{d} x}[-5]\right)(x-1)+(x-5)\left(\frac{\mathrm{d}}{\mathrm{d} x}[x]+\frac{\mathrm{d}}{\mathrm{d} x}[-1]\right)}{2 \sqrt{-(x-5)(x-1)}}\\
&=-\frac{(1+0)(x-1)+(x-5)(1+0)}{2 \sqrt{-(x-5)(x-1)}}\\
&=-\frac{2 x-6}{2 \sqrt{-(x-5)(x-1)}}
\end{aligned}

 

Rewrite/simplify:
\[
=\frac{6-2 x}{2 \sqrt{-(x-5)(x-1)}}
\]
Simplify/rewrite:
\[
-\frac{x-3}{\sqrt{-(x-5)(x-1)}}
\]
Root/zero found at:
\[
x=3
\]

 

Second Derivative

\begin{array}{r}
\frac{\mathrm{d}}{\mathrm{d} x}\left[-\frac{x-3}{\sqrt{-(x-5)(x-1)}}\right] \\
=-\frac{\mathrm{d}}{\mathrm{d} x}\left[\frac{x-3}{\sqrt{-(x-5)(x-1)}}\right] \\
=-\frac{\frac{\mathrm{d}}{\mathrm{d} x}[x-3] \cdot \sqrt{-(x-5)(x-1)}-(x-3) \cdot \frac{\mathrm{d}}{\mathrm{d} x}[\sqrt{-(x-5)(x-1)}]}{(\sqrt{-(x-5)(x-1)})^{2}} \\
=\frac{\left(\frac{\mathrm{d}}{\mathrm{d} x}[x]+\frac{\mathrm{d}}{\mathrm{d} x}[-3]\right) \sqrt{-(x-5)(x-1)}-(x-3) \cdot \frac{1}{2}(-(x-5)(x-1))^{\frac{1}{2}-1} \cdot \frac{\mathrm{d}}{\mathrm{d} x}[-(x-5)(x-1)]}{(x-5)(x-1)}
\end{array}

\begin{aligned}
&=\frac{(1+0) \sqrt{-(x-5)(x-1)}-\frac{(x-3)\left(-\frac{\mathrm{d}}{\mathrm{d} x}[(x-5)(x-1)]\right)}{2 \sqrt{-(x-5)(x-1)}}}{(x-5)(x-1)}\\
&=\frac{\sqrt{-(x-5)(x-1)}+\frac{(x-3)\left(\frac{\mathrm{d}}{\mathrm{d} x}[x-5] \cdot(x-1)+(x-5) \cdot \frac{\mathrm{d}}{\mathrm{d} x}[x-1]\right)}{2 \sqrt{-(x-5)(x-1)}}}{(x-5)(x-1)}\\
&=\frac{\sqrt{-(x-5)(x-1)}+\frac{(x-3)\left(\left(\frac{\mathrm{d}}{\mathrm{d} x}[x]+\frac{\mathrm{d}}{\mathrm{d} x}[-5]\right)(x-1)+(x-5)\left(\frac{\mathrm{d}}{\mathrm{d} x}[x]+\frac{\mathrm{d}}{\mathrm{d} x}[-1]\right)\right)}{2 \sqrt{-(x-5)(x-1)}}}{(x-5)(x-1)}\\
&=\frac{\sqrt{-(x-5)(x-1)}+\frac{(x-3)((1+0)(x-1)+(x-5)(1+0))}{2 \sqrt{-(x-5)(x-1)}}}{(x-5)(x-1)}\\
&=\frac{\frac{(x-3)(2 x-6)}{2 \sqrt{-(x-5)(x-1)}}+\sqrt{-(x-5)(x-1)}}{(x-5)(x-1)}
\end{aligned}

Rewrite/simplify:
\[
=\frac{(6-2 x)(x-3)}{2(-(x-5)(x-1))^{\frac{3}{2}}}-\frac{1}{\sqrt{-(x-5)(x-1)}}
\]
Simplify/rewrite:
\[
-\frac{4}{(-(x-5)(x-1))^{\frac{3}{2}}}
\]

 

 

 

by Gold Status (31,789 points)

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