Maximum value of function f(x)

Maximum value of $f(x) = \left|\sqrt{x^2-8x+52} - \sqrt{x^2-4x+8}\right|$ , where $x\in \mathbb{R}$

#Algebra #MathProblem #Math

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

consider point A moving in x axis and B (4 6) C (2 2), f(x) can be seemd as AB-AC, apply triangle inequality
AB-AC<=BC=2sqrt{5}

Wentao Xu - 7 years, 6 months ago

very nice way !

Piyushkumar Palan - 7 years, 6 months ago

Consider, $\sqrt{x^2 - 8x + 52} \geq \sqrt{x^2-4x+8}$

$\Rightarrow x^2 -8x + 52 \geq x^2 -4x + 8 \Rightarrow x \leq 11$ First Consider $x \leq 11$

$f(x) = \sqrt{x^2 - 8x + 52} \geq \sqrt{x^2-4x+8}$

$\Rightarrow f'(x) = \frac{x-4}{\sqrt{x^2-8x+52}} - \frac{x-2}{\sqrt{x^2-4x+8}}$

Clearly , $f'(x) >0$ $\forall$ $x , 1, f'(x) = 0$ at $x= 1,$ and $f'(x) < 0$ $\forall x$ $\in (1, 11)$ .

Hence, $x=1$ is a local maximum . Also, $f(1) = 2 \sqrt{5}$ .

Now, $\forall x . 11$ , only the sign of the expression changes, hence $x=1$ is the only extrema of the function.

Wait! Don't just say $2 \sqrt{5}$ is the answer, you also have to check $\displaystyle \lim_{x \to \infty} f(x)$ , as $f(x)$ is constantly increasing $\forall x> 11$ , and hence there might come one $x$ when $f(x)$ crosses $2 \sqrt{5}$

To find $\displaystyle \lim_{x \to \infty} f(x)$ ,rationalise the numerator to get,

$f(x) = \frac{4(x-11)}{\sqrt{x^2 - 8x+52} + \sqrt{x^2 - 4x + 8}}$

Divide numerator and denominator by x to get :

$\displaystyle \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{4(1 - \frac{11}{x})}{\sqrt{1 - \frac{8}{x} + \frac{52}{x^2}} + \sqrt{1 - \frac{4}{x} + \frac{8}{x^2}}}$

= $2$ , which is less than $2 \sqrt{5}$

Hence, we conclude that the max. value is $\boxed{2 \sqrt{5}}$

Here is the graph of this function.

jatin yadav - 7 years, 6 months ago

Thanks friends for Nice solution.

I have tried like this way.

$y = f(x) = \left|\sqrt{x^2-8x+52}-\sqrt{x^2-4x+8}\right|\Rightarrow y^2 = \left|\sqrt{x^2-8x+52}-\sqrt{x^2-4x+8}\right|^2$

So $y^2 = \left\{\sqrt{x^2-8x+52}-\sqrt{x^2-4x+8}\right\}^2 = x^2-8x+52+x^2-4x+8-2\sqrt{(x^2-8x+52)\cdot (x^2-4x+8)}$

So $y^2= 2(x^2-6x+30)-2\sqrt{\left\{(x-4)^2+6^2\right\}\cdot \left\{(x-2)^2+2^2\right\}}$

Now Using cauchy - schtwartz Inequality

$\left\{(x-4)^2+6^2\right\}\cdot \left\{(x-2)^2+2^2\right\}\geq \left\{(x-4)\cdot (x-2)+6\cdot 2\right\}^2$

So $\sqrt{\left\{(x-4)^2+6^2\right\}\cdot \left\{(x-2)^2+2^2\right\}}\geq \left\{x^2-6x+20\right\}$

So $-\sqrt{\left\{(x-4)^2+6^2\right\}\cdot \left\{(x-2)^2+2^2\right\}}\leq - \left\{x^2-6x+20\right\}$

So $y^2\leq 2(x^2-6x+30)-2(x^2-6x+20) = 20\Rightarrow y \leq 2\sqrt{5}$ because $y\geq 0$

jagdish singh - 7 years, 6 months ago

Consider the triangle formed by A(x,0), B(2,2) and C(4,6), f(x) = |AC - AB| <= BC = 2sqrt(5). The equality holds when the triangle is degenerated, that is when A, B, C are co-linear. So A is (1,0) or x=1.

George G - 7 years, 6 months ago

Is it 2*sqrt(5).. ?

Rupansh Deshkar - 7 years, 6 months ago

Wolfram|Alpha says $2\sqrt{5}$ at $x=1$

Daniel Leong - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$