what-is-minimum-value-of-function-f-x-x-2-4-x-2-24x-153-

Question Number 77810 by jagoll last updated on 10/Jan/20

w h a t i s m i n i m u m v a l u e

o f f u n c t i o n f (x) =

\sqrt{x^{2} + 4} + \sqrt{x^{2} - 24 x + 153}

Answered by MJS last updated on 10/Jan/20

\sqrt{x^{2} + 4} ⩾ 2, the minimum is at x = 0 \Rightarrow

\Rightarrow \sqrt{x^{2} + 4} is strictly increasing for x \neq 0

x^{2} - 24 x + 1 = 0 \Leftrightarrow x = 12 \pm \sqrt{143} \Rightarrow

\Rightarrow \sqrt{x^{2} - 24 x + 1} is not defined for

12 - \sqrt{143} < x < 12 + \sqrt{143}, especially not definef

for x = 0 \Rightarrow the minimum of \sqrt{x^{2} + 4} + \sqrt{x^{2} - 24 x + 1}

is at one or both of these borders

x = 12 - \sqrt{143} \lor x = 12 + \sqrt{143} \Rightarrow

\Rightarrow minimum is at x = 12 + \sqrt{143} and has the

value \sqrt{291 - 24 \sqrt{143}}

Commented by jagoll last updated on 10/Jan/20

s i r w h y n o t u s e c r i t i c a l p o i n t f r o m

f^{'} (x) = 0 ?

Commented by jagoll last updated on 10/Jan/20

m y t y p o s i r o r i g i n a l p r o b l e m

f (x) = \sqrt{x^{2} + 4} + \sqrt{x^{2} - 24 x + 153}

Commented by MJS last updated on 10/Jan/20

because f^{'} (x) = 0 has got no real solution

f^{'} (x) = \frac{x}{\sqrt{x^{2} + 4}} + \frac{x - 12}{\sqrt{x^{2} - 24 x + 1}} = 0

x \sqrt{x^{2} - 24 x + 1} = (12 - x) \sqrt{x^{2} + 4}

squaring [might leads to false solutions, we

must check them!]

x^{2} (x^{2} - 24 x + 1) = {(12 - x)}^{2} (x^{2} + 4)

\Rightarrow

x^{2} - \frac{32}{49} x + \frac{192}{49} = 0

\Rightarrow x = \frac{16}{49} \pm \frac{8 \sqrt{143}}{49} i

both solve f^{'} (x) = 0 but both are not real

Commented by MJS last updated on 10/Jan/20

anyway I hope you learned something new \dots

Answered by MJS last updated on 10/Jan/20

with the new equation we get

f^{'} (x) = \frac{x}{\sqrt{x^{2} + 4}} + \frac{x - 12}{\sqrt{x^{2} - 24 x + 153}} = 0

x \sqrt{x^{2} - 24 x + 153} = (12 - x) \sqrt{x^{2} + 4}

squaring and transforming

x^{2} + \frac{96}{5} x - \frac{576}{5} = 0

\Rightarrow x = - 24 \lor x = \frac{24}{5}

x = - 24 is false; it {doesn}^{'} t solve f^{'} (x) = 0

\Rightarrow x = \frac{24}{5}

\Rightarrow minmum at x = \frac{24}{5}; f (\frac{24}{5}) = 13

Commented by jagoll last updated on 10/Jan/20

o o y e s t h a n k s y o u s i r

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