Question

\(\large \begin{align} \color{black}{\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}
\hspace{.33em}\\~\\
\normalsize \text{then}
\hspace{.33em}\\~\\
\dfrac{a-d}{b-c}\geq x
\hspace{.33em}\\~\\
\normalsize \text{find x}
\hspace{.33em}\\~\\
a.)2
\hspace{.33em}\\~\\
b.)3
\hspace{.33em}\\~\\
c.)1
\hspace{.33em}\\~\\
d.)0
\hspace{.33em}\\~\\
}\end{align}\)

Answer 1

\(\Large\tt \begin{align} \color{black}{\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k
\hspace{.33em}\\~\\~\\
\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\\~\\
\dfrac{a-d}{b-c}
\hspace{.33em}\\~\\
=\dfrac{bk-\frac{c}{k}}{ck-c}
\hspace{.33em}\\~\\
=\dfrac{(ck)k-\frac{c}{k}}{ck-c}
\hspace{.33em}\\~\\
=\dfrac{k^2-\frac{1}{k}}{k-1}
\hspace{.33em}\\~\\
=\dfrac{k^3-1}{k(k-1)}
\hspace{.33em}\\~\\
=\dfrac{k^3-1}{k(k-1)}
\hspace{.33em}\\~\\
=\dfrac{(k-1)(k^2+k+1)}{k(k-1)}
\hspace{.33em}\\~\\
=1+k+\dfrac{1}{k}
\hspace{.33em}\\~\\
}\end{align}\)

Answer 2

\(\large \begin{align} \color{black}{f(x)=1+x+\dfrac{1}{x}
\hspace{.33em}\\~\\
f(x)'=1-\dfrac{1}{x^2}
\hspace{.33em}\\~\\
0=1-\dfrac{1}{x^2}
\hspace{.33em}\\~\\
x=\pm1
\hspace{.33em}\\~\\
f(x)_{max}=1+1+\dfrac{1}{1}
\hspace{.33em}\\~\\
f(x)_{max}=3
}\end{align}\)

Answer 3

A slightly different way I did it was this: \[\Large a=k^3d \\ \Large b=k^2d \\ \Large c=kd\] Plug them in \[\Large \frac{k^3d-d}{k^2d-kd}=\frac{k^3-1}{k^2-k}= \frac{k^2+k+1}{k}=k+1+\frac{1}{k}\]

Then you can finish just lie you did, or take the cool, completely ridiculous route I took: \[\Large 1+k+\frac{1}{k} = 1+2 \cosh (\ln(k))\] Then take the derivative of this \[\Large \frac{d}{dk} \left( 1+2 \cosh (\ln(k)) \right) =2 \sinh(\ln(k))\frac{1}{k}=0\] So either sinh or 1/k is 0, but we obviously don't want infinity, so we throw away the 1/k part (along with dividing by 2) \[\Large \sinh(\ln(k))=0\] Similar to sin(0)=0, sinh(0)=0 so ln(k) must be 0 as well. \[\Large \ln(k)=0\] And of course, log of 1 is 0, so k=1, just like you got. Just a weird way to the same answer.

Answer 4

looks cool

i dont get why u took

\(\large \begin{align} \color{black}{ 1+k+\dfrac{1}{k} = 1+2 \cosh (\ln(k))
\hspace{.33em}\\~\\

}\end{align}\)

Answer 5

I just wanted to do something weird, the main trick I wanted to show you was how I made everything k^p*d and plugged it in at the start, it made the calculation much cleaner. That second part I was just bored messing around and thought it would be interesting to see. =P

Answer 6

ok , i prefer any method which is short and lazy