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Find the minimum and maximum value of the function $f(x)=3\sqrt{x-2}+\sqrt{4-x}$

My approach is as follows

The domain of the function is $x\in[2,4]$

If the function is $T(x)=\sqrt{x-2}+\sqrt{4-x}$

On squaring we get $(T(x))^2=x-2+4-x+2 \sqrt{(x-2)(4-x)}$

Hence $(T(x))^2=2+2 \sqrt{(x-2)(4-x)}$

Let $(x-2)(4-x)$ represent quadratic equation whose maximum value is $1$ at $x=3$ based on this concept how we will solve it

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  • $\begingroup$ With the substitution $ x = 3+\cos(2t)$, it becomes $ \sqrt{2}(3\cos(t)+\sin(t)=\sqrt{2}\sqrt{10}\cos(2t+a)$ $\endgroup$ Commented Apr 7 at 11:33
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    $\begingroup$ Have you tried setting the derivative equal to zero? $\endgroup$
    – Alex K
    Commented Apr 7 at 11:34
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    $\begingroup$ $x=2 \sin^2 \theta + 4 \cos^2 \theta$ $\endgroup$ Commented Apr 8 at 6:39

3 Answers 3

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You can use the Power-Mean Inequality.

Let $a_1=a_2=...=a_9=\displaystyle{\frac{x-2}{9}}$ and $a_{10}=4-x$. Then $$\displaystyle{\bigg(\frac{f(x)}{10}\bigg)^2=\bigg(\sum\limits_{i=1}^{10}\frac{a_i^{\frac{1}{2}}}{10}\bigg)^2\leq\sum\limits_{i=1}^{10}\frac{a_i}{10}=\frac{1}{5}}$$ Therefore $\displaystyle{f(x)\leq 2\sqrt{5}}$.

Or you can use the Cauchy inequality $$20=((\sqrt{x-2})^2+(\sqrt{4-x})^2)(3^2+1^2)\geq (3\sqrt{x-2}+\sqrt{4-x})^2=f(x)^2$$

On the other hand, let $a=3\sqrt{x-2}$ and $b=\sqrt{4-x}$. Immediately we have $\displaystyle{\frac{a^2}{9}+b^2=2}$, which is an ellipse. Then $\displaystyle{f(x)^2=a^2+b^2+2ab=a^2+2-\frac{a^2}{9}+2ab=\frac{8a^2}{9}+2ab+2\geq 2}$.

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I'm going to write two solutions.

The domain of $f(x)$ is $x\in [2,4]$.

solution 1 :

As commented by Alex K, considering the derivative helps.

We have $$\begin{align}f'(x)&=\frac{3}{2\sqrt{x-2}}-\frac{1}{2\sqrt{4-x}} \\\\&=\frac{3\sqrt{4-x}-\sqrt{x-2}}{2\sqrt{(x-2)(4-x)}} \\\\&=\frac{3\sqrt{4-x}-\sqrt{x-2}}{2\sqrt{(x-2)(4-x)}}\times\frac{3\sqrt{4-x}+\sqrt{x-2}}{3\sqrt{4-x}+\sqrt{x-2}} \\\\&=\frac{9(4-x)-(x-2)}{2\sqrt{(x-2)(4-x)}(3\sqrt{4-x}+\sqrt{x-2})} \\\\&=\frac{-5x+19}{\sqrt{(x-2)(4-x)}(3\sqrt{4-x}+\sqrt{x-2})}\end{align}$$

So, we can see that $f(x)$ is increasing for $x\lt\frac{19}{5}(=3.8)$, and is decreasing for $\frac{19}{5}\lt x$.

Therefore, we see that the maximum value is $$f\bigg(\frac{19}{5}\bigg)=\color{red}{2\sqrt{5}}$$

and that the minimum value is $$\min(f(2),f(4))=f(2)=\color{red}{\sqrt 2}$$


solution 2 :

As commented by Hari Shankar, setting $x=2+2\cos^2 t\ (0\le t\le\frac{\pi}{2})$ works.

We have $$3\sqrt{x-2}+\sqrt{4-x}=3\sqrt 2\cos t+\sqrt 2\sin t$$$$=2\sqrt 5\sin(t+\alpha):=g(t)$$ where $\cos\alpha=\frac{\sqrt 2}{2\sqrt 5}$ and $\sin\alpha=\frac{3\sqrt 2}{2\sqrt 5}$.

So, the maximum value is $g(\frac{\pi}{2}-\alpha)=\color{red}{2\sqrt 5}$.

The minimum value is $$\min\bigg(g(0),g\bigg(\frac{\pi}{2}\bigg)\bigg)=g\bigg(\frac{\pi}{2}\bigg)=\color{red}{\sqrt 2}$$

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Too long for the comment.

  • Domain of $f(x)$ is $[2,4].$

  • The equation $\;f'(x)=0,\;$ or $\,\dfrac3{2\sqrt{x-2}}=\dfrac1{2\sqrt{4-x}},\;$ has the single real root $\;x=\dfrac{19}5.\;$

  • Extremes of $\;f(x)\;$ can be achieved at the edges of domain or at the stationary points, i.e. at the points set $\;x\in\left\{2, \dfrac{19}5,4\right\},\;$ where $\;f(x)\in\{\sqrt2, 2\sqrt5, 3\sqrt2\},\;$ with the least value $\;f\left(2\right)=\sqrt2\;$ and the greatest value $\;f\left(\dfrac{19}5\right)=2\sqrt5.\;$

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