f-x-x-2-4x-13-x-2-14x-130-minimum-value-of-f-x-x-R-

Question Number 175871 by infinityaction last updated on 08/Sep/22

f (x) = \sqrt{x^{2} - 4 x + 13} + \sqrt{x^{2} - 14 x + 130}

minimum value of f (x) x \in R

Answered by mr W last updated on 08/Sep/22

f (x) = \sqrt{{(x - 2)}^{2} + 3^{2}} + \sqrt{{(7 - x)}^{2} + 9^{2}}

⩾ \sqrt{{(x - 2 + 7 - x)}^{2} + {(3 + 9)}^{2}} = \sqrt{5^{2} + 12^{2}} = 13

\Rightarrow f {(x)}_{m i n} = 13

Answered by cortano1 last updated on 08/Sep/22

f (x) = \sqrt{{(2 - x)}^{2} + 9} + \sqrt{{(x - 7)}^{2} + 81}

f (x) \min when \frac{2 - x}{x - 7} = \frac{3}{9}

\frac{2 - x}{x - 7} = \frac{1}{3} \Rightarrow x - 7 = 6 - 3 x

x = \frac{13}{4} \Rightarrow \min f (x) = \sqrt{{(2 - \frac{13}{4})}^{2} + 9} + \sqrt{{(\frac{13}{4} - 7)}^{2} + 81}

= \sqrt{\frac{25}{16} + 9} + \sqrt{\frac{225}{16} + 81}

= \sqrt{\frac{169}{16}} + \sqrt{\frac{1521}{16}} = 13

Commented by infinityaction last updated on 08/Sep/22

explain your 2^{nd} line

this \frac{2 - x}{x - 7} = \frac{3}{9}