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# Maximum value of function f(x)

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

Note by Jagdish Singh
4 years, 1 month ago

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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}

- 4 years, 1 month ago

very nice way !

- 4 years, 1 month 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.

- 4 years, 1 month 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 $$y\geq 0$$

- 4 years, 1 month 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.

- 4 years, 1 month ago

Is it 2*sqrt(5).. ?

- 4 years, 1 month ago

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

- 4 years, 1 month ago