- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,680

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,680

- Admin
- #2

- Jan 26, 2012

- 4,191

$$f(x)= \left( \sqrt{x^{2}-8x+27-6 \sqrt{2}}+ \sqrt{x^{2}-4x+7-2 \sqrt{2}} \right)^{4},$$

and let

\begin{align*}

g(x)&=x^{2}-8x+27-6 \sqrt{2} \\

h(x)&=x^{2}-4x+7-2 \sqrt{2}.

\end{align*}

By completing the square, we can determine that $g(x)>0$ and $h(x)>0$ for all $x \in \mathbb{R}$. Hence, $ \mathcal{D}(f)= \mathbb{R}$.

Since the fourth root function is monotone increasing, it follows that $f(x)$ has a minimum at the same location as $\hat{f}(x)= \sqrt[4]{f(x)}$.

Taking the derivative $\hat{f}'(x)$ and setting it equal to zero yields the equation

$$(x-4) \sqrt{x^{2}-4x+7-2 \sqrt{2}}+(x-2) \sqrt{x^{2}-8x+27-6 \sqrt{2}}=0.$$

Squaring both sides and expanding out yields the equation

$$(4 \sqrt{2}-8)x^{2}+(8 \sqrt{2}-20)x+8 \sqrt{2}-4=0.$$

The solutions are

$$x=1+ \sqrt{2} \quad \text{or} \quad x= \frac{4- \sqrt{2}}{2}.$$

Plugging these values into $\hat{f}$ yield

\begin{align*}

\hat{f}(1+ \sqrt{2})&=2 \sqrt{2} \approx 2.8284 \\

\hat{f}((4- \sqrt{2})/2)&= \frac{ \left((31-8 \sqrt{2}) \sqrt{31+8 \sqrt{2}}

+(49-28 \sqrt{2}) \sqrt{7+4 \sqrt{2}} \right) \sqrt{34}}{2 \cdot 7 \cdot 17}

\approx 3.9569.

\end{align*}

Hence, the minimum of $f$ is $(2 \sqrt{2})^{4}=(2^{3/2})^{4}=64.$

- Thread starter
- Admin
- #3

- Feb 14, 2012

- 3,680

Wow...what a brilliant way to tackle this challenge problem, well done,

$$f(x)= \left( \sqrt{x^{2}-8x+27-6 \sqrt{2}}+ \sqrt{x^{2}-4x+7-2 \sqrt{2}} \right)^{4},$$

and let

\begin{align*}

g(x)&=x^{2}-8x+27-6 \sqrt{2} \\

h(x)&=x^{2}-4x+7-2 \sqrt{2}.

\end{align*}

By completing the square, we can determine that $g(x)>0$ and $h(x)>0$ for all $x \in \mathbb{R}$. Hence, $ \mathcal{D}(f)= \mathbb{R}$.

Since the fourth root function is monotone increasing, it follows that $f(x)$ has a minimum at the same location as $\hat{f}(x)= \sqrt[4]{f(x)}$.

Taking the derivative $\hat{f}'(x)$ and setting it equal to zero yields the equation

$$(x-4) \sqrt{x^{2}-4x+7-2 \sqrt{2}}+(x-2) \sqrt{x^{2}-8x+27-6 \sqrt{2}}=0.$$

Squaring both sides and expanding out yields the equation

$$(4 \sqrt{2}-8)x^{2}+(8 \sqrt{2}-20)x+8 \sqrt{2}-4=0.$$

The solutions are

$$x=1+ \sqrt{2} \quad \text{or} \quad x= \frac{4- \sqrt{2}}{2}.$$

Plugging these values into $\hat{f}$ yield

\begin{align*}

\hat{f}(1+ \sqrt{2})&=2 \sqrt{2} \approx 2.8284 \\

\hat{f}((4- \sqrt{2})/2)&= \frac{ \left((31-8 \sqrt{2}) \sqrt{31+8 \sqrt{2}}

+(49-28 \sqrt{2}) \sqrt{7+4 \sqrt{2}} \right) \sqrt{34}}{2 \cdot 7 \cdot 17}

\approx 3.9569.

\end{align*}

Hence, the minimum of $f$ is $(2 \sqrt{2})^{4}=(2^{3/2})^{4}=64.$

I learned something valuable from your method, thank you for your solution and thank you for participating!