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,

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.

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TopNewestconsider 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}

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very nice way !

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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.

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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 \(y\geq 0\)

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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.

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Is it 2*sqrt(5).. ?

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

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