min-f-x-x-2-2x-5-4x-2-4x-10-

Question Number 175623 by cortano1 last updated on 04/Sep/22

\min f (x) = \sqrt{x^{2} - 2 x + 5} + \sqrt{{4 x}^{2} - 4 x + 10}

Answered by greougoury555 last updated on 04/Sep/22

l e t {\begin{cases} a = (1 - x) i + 2 j \\ b = (2 x - 1) i + 3 j \end{cases}

∣ a ∣ + ∣ b ∣ ⩾ ∣ a + b ∣

f (x) ⩾ \sqrt{x^{2} + 5^{2}}

h o l d s w h e n \frac{1 - x}{2 x - 1} = \frac{2}{3}

4 x - 2 = 3 - 3 x \Rightarrow x = \frac{5}{7}

M i n f (x) = \sqrt{\frac{25}{49} + 25} = \frac{25 \sqrt{2}}{7}

Commented by infinityaction last updated on 04/Sep/22

w h a t i s i a n d j ? ?

Commented by mr W last updated on 04/Sep/22

i a n d j a r e u n i t v e c t o r s i n t h e p l a n e .

Commented by mr W last updated on 04/Sep/22

i t h i n k t h e s o l u t i o n i s w r o n g .

f r o m f (x) ⩾ \sqrt{x^{2} + 5^{2}}

y o u c a n n o t g e t t h e m i n i m u m o f f (x),

b e c a u s e \sqrt{x^{2} + 5^{2}} i s n o t a c o n s t a n t,

b u t a f u n c t i o n w h i c h i n t u r n h a s i t s

o n w m i n i m u m . y o u c a n n o t s a y

a t x = \frac{5}{7} \sqrt{x^{2} + 5^{2}} i s m i n i m u m .

Commented by infinityaction last updated on 04/Sep/22

t h a n k s s i r